Class 12 RD Sharma Solutions- Chapter 28 The Straight Line in Space – Exercise 28.4

Read

Discuss

Question 1. Find the perpendicular distance of the point (3, -1, 11) from the line $\frac{x}{2}=\frac{y-2}{-3}=\frac{z-3}{4}$

Solution:

Let the foot of the perpendicular drawn from P (3, -1, 11) to the line $\frac{x}{2}=\frac{y-2}{-3}=\frac{z-3}{4}$ is Q, so we have to find length of PQ is general point on the line $\frac{x}{2}=\frac{y-2}{-3}=\frac{z-3}{4}=\lambda (say)$

Coordinate of Q = $(2\lambda,-3\lambda+2,4\lambda+3)$ , direction ratios of the given line = 2,-3,4. Since PQ is the perpendicular to the given line interface.

$a_1a_2+b_1b_2+c_1c_2=0$

$\Rightarrow 2(2\lambda-3)+(-3 )(-3\lambda+3)4(4\lambda-8)=0$

$\Rightarrow 4\lambda-6+9\lambda-9+16\lambda-32=0$

$29\lambda-47=0; \lambda = \frac{47}{29}$

So, the coordinates of Q are:

$\Rightarrow 2(\frac{47}{29}),-3(\frac{47}{29})+2,4(\frac{47}{29})+3 = \frac{94}{29},\frac{-83}{29},\frac{275}{29}$

Distance between P and Q is given as:

$= \sqrt{(\frac{94}{29}-3)^2+(\frac{-83}{29}+1)^2+(\frac{275}{29}-11)^2}$

$= \sqrt{(\frac{7}{29})^2+(\frac{-54}{29})^2+(\frac{-44}{29})^2}$

$=\sqrt{\frac{49}{841}+\frac{2916}{841}+\frac{1936}{841}}$

$=\sqrt{\frac{4901}{841}}$

So, the required distance is $\sqrt{\frac{4901}{841}}$ units

Question 2. Find the perpendicular distance of the point (1,0,0) from the line $\frac{x-1}{2}=\frac{y+1}{-3}=\frac{z+10}{8}$ . Also, find the coordinates of the foot of the perpendicular and the equation of the perpendicular.

Solution:

Let us consider the foot of the perpendicular drawn from P (1,0,0) to the line $\frac{x-1}{2}=\frac{y+1}{-3}=\frac{z+10}{8}$ is Q. So let us find the length of PQ i.e. $\frac{x-1}{2}=\frac{y+1}{-3}=\frac{z+10}{8}=\lambda$

Coordinate of Q = $(2\lambda+1, -3\lambda-1, 8\lambda-10)$

The direction ratios of the given line: $a_1a_2+b_1b_2+c_1c_2=0$

$\Rightarrow 2(2\lambda)+(-3)(-3\lambda-1)+8(8\lambda-10)=0$

$4\lambda+9\lambda+3+64\lambda-80=0$

$\Rightarrow 77\lambda-77=0; \lambda = 1$

So the Coordinates of Q are as follows:

$(2\lambda+1, -3\lambda-1, 8\lambda-10)= [2(1)+1,-3(1)-1,8(1)-10]= [3,4,-2]$

Distance between P and Q is given by:

PQ = $\sqrt{(x1-x_2)^2+(y_1-y_2)^2+(z_1-z_2)^2}$

PQ = $\sqrt{(1-3)^2+(0+4)^2+(0+2)^2}=\sqrt{4+16+4}$

PQ = $\sqrt{24} = 2\sqrt{6}$

Hence, the foot of the perpendicular = (3,-4,-2);

Length of the perpendicular = $2\sqrt{6}$ units.

Question 3. Find the foot of the perpendicular drawn from the point A(1,0,3) to the joint of the points B(4,7,1) and C(3,5,3).

Solution:

Let us consider, the foot of the perpendicular drawn from A(1,0,3) to the line joining

Points B(4,7,1) and C(3,5,3) be D. The equation of the line passing through

points B(4,7,1) and C(3,5,3) is

$\frac{x-x_1}{x_2-x_1}=\frac{y-y_1}{y_2-y_1}=\frac{z-z_1}{z_2-z_1}$

$\Rightarrow \frac{x-4}{3-4}=\frac{y-7}{5-7}=\frac{z-1}{3-1}$

$\Rightarrow \frac{x-4}{-1}=\frac{y-7}{-2}=\frac{z-1}{2}$

Let $\frac{x-4}{-1}=\frac{y-7}{-2}=\frac{z-1}{2}=\lambda$

So, the direction ratio of AD is $(-\lambda+4-1),(-2\lambda+7-0),(2\lambda+1-3) = (-\lambda+3),(-2\lambda+7),(2\lambda-2)$

Line AD is the perpendicular to BC so, $a_1a_2+b_1b_2+c_1c_2=0$

$\Rightarrow -1(-\lambda+3)+(2)(-2\lambda+7)+2(2\lambda-2)=0$

$\Rightarrow 9\lambda -21 = 0; \lambda = \frac{21}{9}$

Hence, coordinates of D are: $\left(\frac{-21}{9}+4,(-2)(\frac{21}{9}+7),2(\frac{21}{9}+1)\right)$

= $\left(\frac{15}{9},\frac{21}{9},\frac{51}{9}\right) = \left(\frac{5}{3},\frac{7}{3},\frac{17}{3}\right)$

Question 4. A (1,0,4), B (0,-11,3), C (2,-3,1) are three points and D is the foot of the perpendicular from A on BC. Find the coordinates of D.

Solution:

Given: D is the perpendicular from A(1,0,4) on BC. So,

Equation of line passing through BC is:

$\frac{x-x_1}{x_2-x_1}=\frac{y-y_1}{y_2-y_1}=\frac{z-z_1}{z_2-z_1}$

$\frac{x-0}{2-0}=\frac{y+11}{-3+11}=\frac{z-3}{1-3}$

$\frac{x}{2}=\frac{y+11}{8}=\frac{z-3}{-2}$

$\frac{x}{2}=\frac{y+11}{8}=\frac{z-3}{-2}=\lambda$

Coordinates of D = ( $2\lambda, 8\lambda-11,-2\lambda+3$ )

Direction ratios of AD is $(2\lambda-1),(8\lambda-11-0),(-2\lambda+3-4)= (2\lambda-1),(8\lambda-11),(-2\lambda-1)$

Line AD is perpendicular to BS so,

$a_1a_2+b_1b_2+c_1c_2=0$

$2(2\lambda-1)+8(8\lambda-11)+(-2)(-2\lambda-1)=0$

$72\lambda-88 = 0; \lambda = \frac{88}{72} or \frac{11}{9}$

So, coordinates of D are = $(2\lambda,8\lambda-11,-2\lambda+3)$

$\left(2(\frac{11}{9}),8(\frac{11}{9})-11,-2(\frac{11}{9})+3\right)$ =

$\left(\frac{22}{9},\frac{-11}{9},\frac{5}{9}\right)$

Question 5. Find the foot of the perpendicular from the point (2,3,4) to the line $\frac{4-x}{2}=\frac{y}{6}=\frac{1-z}{3}$ . Also, find the perpendicular distance from the given point to the line.

Solution:

Let us consider that The foot of the perpendicular drawn from P(2,3,4) to the line

$\frac{4-x}{2}=\frac{y}{6}=\frac{1-z}{3}$ is $\theta$ .

Equation of the line is $\frac{4-x}{2}=\frac{y}{6}=\frac{1-z}{3}$

Let $\frac{x-4}{-2}=\frac{y}{6}=\frac{z-1}{-3} = \lambda$

Coordinates of Q = $(-2\lambda+4-2),(6\lambda-3),(-3\lambda+1-4) = (-2\lambda+2),(6\lambda-3),(-3\lambda-3)$

So, PQ is perpendicular to the given line, $a_1a_2+b_1b_2+c_1c_2=0$

$(-2)(-2\lambda+2)+6(6\lambda-3)+(-3)(-3\lambda-3)=0$

$49\lambda-13=0; \lambda=\frac{13}{49}$

Coordinates of Q = $(-2\lambda+4, 6\lambda. -3\lambda+1)$

= $\left(-2(\frac{13}{49})+4,6(\frac{13}{49}),-3(\frac{13}{49})+1\right)$

= $\left(\frac{-26+196}{49},\frac{78}{49},\frac{-39+49}{49}\right)= \left(\frac{170}{49}, \frac{78}{49},\frac{10}{49}\right)$

Distance between P and Q is given by: PQ = $\sqrt{(x1-x_2)^2+(y_1-y_2)^2+(z_1-z_2)^2}$

= $\sqrt{(\frac{170}{49}-2)^2+(\frac{78}{49}-3)^2+(\frac{10}{49}-4)^2}$

= $\sqrt{\frac{44541}{2401}} = \sqrt{\frac{909}{49}} = \frac{3\sqrt{101}}{49}$

Hence, perpendicular distance from (2,3,4) to the given line is $\frac{3\sqrt{101}}{49}$ units.

Question 6. Find the equation of the perpendicular drawn from the point P (2,4,-1) to the line $\frac{x+5}{1}=\frac{y+3}{4}=\frac{z-6}{-9}$ . Also, write down the coordinates of the foot of the perpendicular from P.

Solution:

Let $\theta$ be the foot of the perpendicular drawn from P(2,4,-1) to the line

$\frac{x+5}{1}=\frac{y+3}{4}=\frac{z-6}{-9}$

Given line is $\frac{x+5}{1}=\frac{y+3}{4}=\frac{z-6}{-9} = \lambda (say)$

Coordinate of Q (General point on the line) = $(\lambda-5, 4\lambda-3, -9\lambda+6)$

Direction ratios of PQ are: $(\lambda-5-2),(4\lambda-3-4),(-9\lambda+6+1) = \lambda-7, 4\lambda-7, -9\lambda+7$

As line PQ is perpendicular to the given line, so: $a_1a_2+b_1b_2+c_1c_2=0$

$1(\lambda-7)+4(4\lambda-7)+(-9)(-9\lambda+7)=0$

$98\lambda-98=0; \lambda=1$

Therefore, coordinates of foot of perpendicular = {-4, 1, -3}

So equation of the perpendicular PQ is : $\frac{x-x_1}{x_2-x_1}=\frac{y-y_1}{y_2-y_1}=\frac{z-z_1}{z_2-z_1}$

$\Rightarrow \frac{x-2}{-4-2}=\frac{y-4}{1-4}=\frac{z+1}{-3+1}$

$\Rightarrow \frac{x-2}{-6}=\frac{y-4}{-3}=\frac{z+1}{-2}$

Question 7. Find the length of the perpendicular drawn from the point (5,4,-1) to the line $\vec{r}=\widehat{i} + \lambda(2\widehat{i} + 9\widehat{j}+5\widehat{k})$

Solution:

Let the foot of the perpendicular drawn from P(5,4,-1) to the given line is Q, so given equation of line is:

$\vec{r}=\widehat{i} + \lambda(2\widehat{i} + 9\widehat{j}+5\widehat{k})$

$(x\widehat{i}+y\widehat{j}+z\widehat{k}) = (1+2\lambda)\widehat{i}+(9\lambda)\widehat{j}+(5\lambda)\widehat{k}$

Equating the coefficients of $\widehat{i}, \widehat{j}, \widehat{k}$

$\Rightarrow x = 1+2\lambda, y = 9\lambda, z = 5\lambda$

$\Rightarrow \frac{x-1}{2} = \lambda, \frac{y}{9} = \lambda, \frac{z}{5}=\lambda$

$\Rightarrow \frac{x-1}{2} = \frac{y}{9} =\frac{z}{5}=\lambda (say)$

Coordinate of Q = $(2\lambda+1,9\lambda,5\lambda)$

Direction ratios of line PQ are: $(2\lambda+1-5),9\lambda-4,5\lambda+1$

$\Rightarrow 2\lambda-4, 9\lambda-4, 5\lambda+1$

As line PQ is perpendicular to the given line, so: $a_1a_2+b_1b_2+c_1c_2=0$

$2(2\lambda-4)+9(9\lambda-4)+5(5\lambda+1)=0$

$\Rightarrow 4\lambda - 8 +81\lambda - 36 +25\lambda + 5 = 0$

$\Rightarrow 110\lambda - 39 = 0; \lambda = \frac{39}{110}$

Coordinate of Q = { $2\lambda+1, 9\lambda, 5\lambda$ }

= $\left(2(\frac{39}{110})+1,9(\frac{39}{110}),5(\frac{39}{110})\right)$

= $\left(\frac{188}{110},\frac{351}{110},\frac{195}{110}\right)$

Length of perpendicular = PQ = $\sqrt{(x1-x_2)^2+(y_1-y_2)^2+(z_1-z_2)^2}$

= $\sqrt{(5-\frac{188}{110})^2+ (4-\frac{351}{110})^2+(-1-\frac{195}{110})^2}$

PQ = $\sqrt{\frac{231990}{12100}} or \sqrt{\frac{2109}{110}}$

Question 8. Find the foot of the perpendicular drawn from the point $\widehat{i}+6\widehat{j}+3\widehat{k}$ to the line $\vec{r} = (\widehat{j}+2\widehat{k}) + \lambda(\widehat{i}+2\widehat{j}+3\widehat{k})$ . Also, find the length of the perpendicular.

Solution:

Let position vector of foot of perpendicular drawn from p $(\widehat{i}+6\widehat{j}+3\widehat{k})$ on $\vec{r} = (\widehat{j}+2\widehat{k}) + \lambda(\widehat{i}+2\widehat{j}+3\widehat{k})$ be Q $(\vec{q})$ . So, Q is on the line $\vec{r} = (\widehat{j}+2\widehat{k}) + \lambda(\widehat{i}+2\widehat{j}+3\widehat{k})$

So, position vector of Q = $\lambda(\widehat{i})+(1+2\lambda)\widehat{j}+(2+3\lambda)\widehat{k}$

$\vec{PQ}$ is the position vector of Q – position vector of p = $(\lambda\widehat{i}+(1+2\lambda)\widehat{j}-(\widehat{i}+6\widehat{j}+3\widehat{k})$

$\vec{PQ} = (\lambda-1)\widehat{i}+(2\lambda-5)\widehat{j}+(3\lambda-1)\widehat{k}$

Here, PQ vector is perpendicular to the given line. So,

$(\lambda-1)\widehat{i}+(2\lambda-5)\widehat{j}+(3\lambda-1)\widehat{k} = 0$

$\Rightarrow (\lambda-1)1+(2\lambda-5)2+(3\lambda-1)3 = 0$

$14\lambda -14 = 0; \lambda = 1$

Position vector of Q = { $(\widehat{j}+2\widehat{j})+\lambda(\widehat{i}+2\widehat{j}+3\widehat{k})$ }

= $(\widehat{j}+2\widehat{k})+1(\widehat{i}+2\widehat{j}+3\widehat{k})$

Foot of the perpendicular = $\widehat{i}+3\widehat{j}+5\widehat{k}$

$\vec{PQ}$ = Position vector of Q – Position vector of P

= $(\widehat{i}+3\widehat{j}+5\widehat{k})-(\widehat{i}+6\widehat{j}+3\widehat{k})$

= $-3\widehat{i}+2\widehat{k}$

$|\vec{PQ}|$ = $\sqrt{(-3)^2+(2)^2} = \sqrt{13}$ units

Question 9. Find the equation of the peprendicular drwan from the point P (-1,3,2) to the line $\vec{r} = (2\widehat{j}+3\widehat{k})+\lambda(2\widehat{i}+\widehat{j}+3\widehat{k})$ . Also, find the coordinates of the foot of the perpendicular from P.

Solution:

Let Q be the perpendicular drawn from P { $-\widehat{i}+3\widehat{j}+2\widehat{k}$ } on the

vector $\vec{r} = (2\widehat{j}+3\widehat{k})+\lambda(2\widehat{i}+\widehat{j}+3\widehat{k})$

Let the position vector of Q be : $(2\widehat{i}+3\widehat{k})+ \lambda(2\widehat{i}+\widehat{j}+3\widehat{k})$

: $(2\lambda)\widehat{i}+(2+\lambda)\widehat{j}+(3+3\lambda)\widehat{k}$

$\vec{PQ}$ = Position Vector of Q – Position Vector of P = $(2\lambda)\widehat{i}+(2+\lambda)\widehat{j}+(3+3\lambda)\widehat{k} - (-\widehat{i}+3\widehat{j}+2\widehat{k})$

$\vec{PQ} = (2\lambda+1)\widehat{i} + (\lambda-1)\widehat{j}+(3\lambda+1)\widehat{k}$

As PQ vector is perpendicular to the given line,

$(2\lambda+1)\widehat{i} + (\lambda-1)\widehat{j}+(3\lambda+1)\widehat{k} = 0$

$4\lambda+2+\lambda-1+9\lambda+3 = 0; \lambda= \frac{-4}{14} or \frac{-2}{7}$

Position Vector of Q = $(2\lambda)\widehat{i}+(2+\lambda)\widehat{j}+(3+3\lambda)\widehat{k}$ is

$2(\frac{-2}{7})\widehat{i}+(2-\frac{2}{7})\widehat{j}+(3+3(\frac{-2}{7})\widehat{k})$

$\frac{-4}{7}\widehat{i}+\frac{12}{7}\widehat{j}+\frac{15}{7}\widehat{k}$

Coordinates of foot of the perpendicular: $\left(\frac{-4}{7},\frac{12}{7},\frac{15}{7}\right)$

Equation of PQ is: $\vec{r} = \vec{a}+\lambda(\vec{b}-\vec{a})$

$\Rightarrow \vec{r} = (-\widehat{i}+3\widehat{j}+2\widehat{k})+\lambda((\frac{-4}{7}\widehat{i}+\frac{12}{7}\widehat{j}+\frac{15}{7}\widehat{k})-(-\widehat{i}+3\widehat{j}+2\widehat{k}))$

$\vec{r} = (-\widehat{i}+3\widehat{j}+2\widehat{k}) + \lambda(\frac{3}{7}\widehat{i}-\frac{9}{7}\widehat{j}+\frac{1}{7}\widehat{k})$

$\vec{r} = (-\widehat{i}+3\widehat{j}+2\widehat{k}) + \mu(3\widehat{i}-9\widehat{j}+\widehat{k})$

Question 10. Find the foot of the perpendicular from (0,2,7) on the line $\frac{x+2}{-1}=\frac{y-1}{3}=\frac{z-3}{-2}$

Solution:

Let the foot of the perpendicular drawn from (0,2,7) to the line $\frac{x+2}{-1}=\frac{y-1}{3}=\frac{z-3}{-2}$ be Q.

Given equation of the line is $\frac{x+2}{-1}=\frac{y-1}{3}=\frac{z-3}{-2} = \lambda (say)$

Coordinate of Q = { $-\lambda-2, 3\lambda+1, -2\lambda+3$ }

Direction Ratios of PQ are $(\lambda-2-0), (3\lambda+1-2), (-2\lambda+3-7) = (-\lambda-2),(3\lambda-1),(-2\lambda- 4)$

Since, PQ is perpendicular to the given line, so $a_1a_2+b_1b_2+c_1c_2=0$

$-1(-\lambda-2)+3(3\lambda-1)+(-2)(-2\lambda-4)=0$

$\Rightarrow \lambda+2+9\lambda-3+4\lambda+8=0$

$\Rightarrow 14\lambda + 7 =0; \lambda = \frac{-1}{2}$

Foot of the perpendicular = { $\lambda-2, 3\lambda+1, -2\lambda+3$ }\

= $\left(-(\frac{-1}{2})-2, 3(\frac{-1}{2}),-2(\frac{-1}{2})+3\right)$

$\left(\frac{-3}{2}, \frac{-1}{2}, 4\right)$

Question 11. Find the foot of the perpendicular from (1,2,-3) to the line $\frac{x+1}{2}=\frac{y-3}{-2}=\frac{z}{-1}$

Solution:

Let the foot of perpendicular from P (1,2,-3) to the line $\frac{x+1}{2}=\frac{y-3}{-2}=\frac{z}{-1}$ be Q.

Given the equation of line is $\frac{x+1}{2}=\frac{y-3}{-2}=\frac{z}{-1} =\lambda (say)$

$\Rightarrow x = 2\lambda-1, y=-2\lambda+3, z = -\lambda$

Coordinates of Q are { $2\lambda-1, -2\lambda+3, -\lambda$ }

Direction Ratios of PQ are: $(2\lambda-1-1), (-2\lambda+3-2), (-\lambda+3)$ =

$(2\lambda-2), (-2\lambda+1), (-\lambda+3)$

Let PQ be the perpendicular to the given line, so $a_1a_2+b_1b_2+c_1c_2=0$

$2(2\lambda-2)+(-2)(-2\lambda+1)+(-1)(-\lambda+3)=0$

$9\lambda-9 = 0; \lambda = 1$

Coordinate of the perpendicular: $(2\lambda-1, -2\lambda+3, -\lambda)$

$\Rightarrow (2(1)-1, -2(1)+3, -1) = (1,1,-1)$

Question 12. Find the equation of the line passing through the points A (0,6,-9) and B (-3, 6, 3). If D is the foot of the perpendicular drawn from a point C (7,4,-1) on the line AB, then find the coordinates of the point D and the equation of the line CD.

Solution:

Equation of line AB is $\frac{x-x_1}{x_2-x_1}=\frac{y-y_1}{y_2-y_1}=\frac{z-z_1}{z_2-z_1}$

$\Rightarrow \frac{x-0}{3-0}=\frac{y-6}{-6-6}=\frac{z+9}{3+9}$

$\Rightarrow \frac{x-0}{3-0}=\frac{y-6}{-6-6}=\frac{z+9}{3+9} = \lambda (say)$

Coordinate of point D = { $-3\lambda, -12\lambda+6, 12\lambda-9$ }

Direction ratios of CD = $(-3\lambda-7)+(-12)(-12\lambda+6-4)+12(12\lambda-9+1)$

= $(-3\lambda-7)+(-12)(-12\lambda+2)+12(12\lambda-8)$

As line CD is perpendicular to the line AB, so $a_1a_2+b_1b_2+c_1c_2=0$

$-3(-3\lambda-7)+(-12)(-12\lambda+2)+12(12\lambda-8) =0$

$297\lambda-99=0; \lambda = \frac{1}{3}$

Coordinate of D = { $-3\lambda, -12\lambda+6, 12\lambda-9$ }

= { $-3(\frac{1}{3}), -12(\frac{1}{3})+6, 12(\frac{1}{3})-9$ }

= (-1,2,-5)

Equation of CD is $\frac{x-x_1}{x_2-x_1}=\frac{y-y_1}{y_2-y_1}=\frac{z-z_1}{z_2-z_1}$

$\Rightarrow \frac{x-7}{-1-7}=\frac{y-4}{2-4}=\frac{z+1}{-5+1}$

$\Rightarrow \frac{x-7}{-8}=\frac{y-4}{-2}=\frac{z+1}{-4}$

$\Rightarrow \frac{x-7}{4}=\frac{y-4}{1}=\frac{z+1}{2}$

Question 13. Find the distance of the point (2,4,-1) from the line $\frac{x+5}{1}=\frac{y+3}{4}=\frac{z-6}{-9}$

Solution:

Let P = (2,4,-1)

In order to find the distance we need to find a point Q on the line. We see that line is passing through

the point Q(-5,-3,6). So, let’s take this point as the required point.

The line is also parallel to the vector $\vec{b} = \widehat{i} +4\widehat{j} -9\widehat{k}$

Now, $\vec{PQ}$ = $(-5\widehat{i}-3\widehat{j}+6\widehat{k})-(2\widehat{i}+4\widehat{j}-\widehat{k})=-7\widehat{i}-7\widehat{j}+7\widehat{k}$

$\vec{b} \times \vec{PQ} = \begin{vmatrix} \widehat{i} & \widehat{j} & \widehat{k} \\ 1 & 4 & -9 \\ -7 & -7 & 7 \end{vmatrix}= -35\widehat{i}+56\widehat{j}+21\widehat{k}$

$|\vec{b} \times \vec{PQ}| = \sqrt{1225+3136+441} = \sqrt{4802}$

$\vec{b} = \sqrt{1+16+81} = \sqrt{98}$

Therefore, $d= \frac{|\vec{b} \times \vec{PQ}|}{|\vec{b}|} = \frac{\sqrt{4802}}{\sqrt{98}}=7$

Question 14. Find the coordinates of the foot of the perpendicular drawn from point A (1,8,4) to the line joining the points B (0,-1,3) and C (2,-3,-1).

Solution:

Let L be the foot of the perpendicular drawn from A(1,8,4) on the line joining the points B(0,-1,3) and C(2,-3,-1).

Equation of the line passing through the points B and C is given by

$\vec{r}=\vec{b}+\lambda(\vec{c}-\vec{b})$

$\vec{r} = (0+2\lambda)\widehat{i}+(-1-2\lambda)\widehat{j}+(3-4\lambda)\widehat{k}$

Let position vector of L be,

$\vec{r}=(2\lambda)\widehat{i}+(-1-2\lambda)\widehat{k}+(3-4\lambda)\widehat{k}......(Equation 1)$

Then, $\vec{AL}$ = Position vector of L – Position vector of A

$\Rightarrow \vec{AL} = (2\lambda)\widehat{i}+(-1-2\lambda)\widehat{j}+(3-4\lambda)\widehat{k}-(1\widehat{i}+8\widehat{j}+4\widehat{k})$

Since, AL vector is perpendicular to the given line

which is parallel to $\vec{b}=2\widehat{i}-2\widehat{j}-4\widehat{k}$

Therefore, $\vec{AL}\cdot\vec{b}=0$

$\Rightarrow 2(-1+2\lambda)-2(-9-2\lambda)-4(-1-4\lambda)=0$

$24\lambda+20=0; \lambda=\frac{-5}{6}$

Putting value of lambda in Equation 1, we get:

$\vec{r} = \frac{-5}{3}\widehat{i}+\frac{2}{3}\widehat{j}+\frac{19}{3}\widehat{k}$

So, coordinates of foot of the perpendicular are

$\left(\frac{-5}{3}, \frac{2}{3}, \frac{19}{3}\right)$

Whether you're preparing for your first job interview or aiming to upskill in this ever-evolving tech landscape, GeeksforGeeks Courses are your key to success. We provide top-quality content at affordable prices, all geared towards accelerating your growth in a time-bound manner. Join the millions we've already empowered, and we're here to do the same for you. Don't miss out - check it out now!

Last Updated : 14 Nov, 2022

Class 12 RD Sharma Solutions - Chapter 28 The Straight Line in Space - Exercise 28.3

Class 12 RD Sharma Solutions - Chapter 28 The Straight Line in Space - Exercise 28.5

Chapter 1: Relations

Chapter 2: Functions

Chapter 3: Binary Operations

Chapter 4: Inverse Trigonometric Functions

Chapter 5: Algebra of Matrices

Chapter 6: Determinants

Chapter 7: Adjoint and Inverse of a Matrix

Chapter 8: Solutions of Simultaneous Linear Equations

Chapter 9: Continuity

Chapter 10: Differentiability

Chapter 11: Differentiation

Chapter 12: Higher Order Derivatives

Chapter 13: Derivative as a Rate Measurer

Chapter 14: Differentials, Errors, and Approximations

Chapter 15: Mean Value Theorems

Chapter 16: Tangents and Normals

Chapter 17: Increasing and Decreasing Functions

Chapter 18: Maxima and Minima

Chapter 19: Indefinite Integrals

Chapter 20: Definite Integrals

Chapter 21: Areas of Bounded Regions

Chapter 22: Differential Equations

Chapter 23: Algebra of Vectors

Chapter 24: Scalar Or Dot Product

Chapter 25: Vector or Cross Product

Chapter 26: Scalar Triple Product

Chapter 27: Direction Cosines and Direction Ratios

Chapter 28: Straight Line in Space

Chapter 29: The Plane

Chapter 30: Linear programming

Chapter 31: Probability

Chapter 32: Mean and Variance of a Random Variable

Chapter 33: Binomial Distribution

Chapter 1: Relations

Chapter 2: Functions

Chapter 3: Binary Operations

Chapter 4: Inverse Trigonometric Functions

Chapter 5: Algebra of Matrices

Chapter 6: Determinants

Chapter 7: Adjoint and Inverse of a Matrix

Chapter 8: Solutions of Simultaneous Linear Equations

Chapter 9: Continuity

Chapter 10: Differentiability

Chapter 11: Differentiation

Chapter 12: Higher Order Derivatives

Chapter 13: Derivative as a Rate Measurer

Chapter 14: Differentials, Errors, and Approximations

Chapter 15: Mean Value Theorems

Chapter 16: Tangents and Normals

Chapter 17: Increasing and Decreasing Functions

Chapter 18: Maxima and Minima

Chapter 19: Indefinite Integrals

Chapter 20: Definite Integrals

Chapter 21: Areas of Bounded Regions

Chapter 22: Differential Equations

Chapter 23: Algebra of Vectors

Chapter 24: Scalar Or Dot Product

Chapter 25: Vector or Cross Product

Chapter 26: Scalar Triple Product

Chapter 27: Direction Cosines and Direction Ratios

Chapter 28: Straight Line in Space

Chapter 29: The Plane

Chapter 30: Linear programming

Chapter 31: Probability

Chapter 32: Mean and Variance of a Random Variable

Chapter 33: Binomial Distribution

Class 12 RD Sharma Solutions- Chapter 28 The Straight Line in Space – Exercise 28.4

Question 1. Find the perpendicular distance of the point (3, -1, 11) from the line

Question 2. Find the perpendicular distance of the point (1,0,0) from the line . Also, find the coordinates of the foot of the perpendicular and the equation of the perpendicular.

Question 3. Find the foot of the perpendicular drawn from the point A(1,0,3) to the joint of the points B(4,7,1) and C(3,5,3).

Question 4. A (1,0,4), B (0,-11,3), C (2,-3,1) are three points and D is the foot of the perpendicular from A on BC. Find the coordinates of D.

Question 5. Find the foot of the perpendicular from the point (2,3,4) to the line . Also, find the perpendicular distance from the given point to the line.

Question 6. Find the equation of the perpendicular drawn from the point P (2,4,-1) to the line . Also, write down the coordinates of the foot of the perpendicular from P.

Question 7. Find the length of the perpendicular drawn from the point (5,4,-1) to the line

Question 8. Find the foot of the perpendicular drawn from the point to the line . Also, find the length of the perpendicular.

Question 9. Find the equation of the peprendicular drwan from the point P (-1,3,2) to the line . Also, find the coordinates of the foot of the perpendicular from P.

Question 10. Find the foot of the perpendicular from (0,2,7) on the line

Question 11. Find the foot of the perpendicular from (1,2,-3) to the line

Question 12. Find the equation of the line passing through the points A (0,6,-9) and B (-3, 6, 3). If D is the foot of the perpendicular drawn from a point C (7,4,-1) on the line AB, then find the coordinates of the point D and the equation of the line CD.

Question 13. Find the distance of the point (2,4,-1) from the line

Question 1. Find the perpendicular distance of the point (3, -1, 11) from the line $\frac{x}{2}=\frac{y-2}{-3}=\frac{z-3}{4}$

Question 2. Find the perpendicular distance of the point (1,0,0) from the line $\frac{x-1}{2}=\frac{y+1}{-3}=\frac{z+10}{8}$ . Also, find the coordinates of the foot of the perpendicular and the equation of the perpendicular.

Question 5. Find the foot of the perpendicular from the point (2,3,4) to the line $\frac{4-x}{2}=\frac{y}{6}=\frac{1-z}{3}$ . Also, find the perpendicular distance from the given point to the line.

Question 6. Find the equation of the perpendicular drawn from the point P (2,4,-1) to the line $\frac{x+5}{1}=\frac{y+3}{4}=\frac{z-6}{-9}$ . Also, write down the coordinates of the foot of the perpendicular from P.

Question 7. Find the length of the perpendicular drawn from the point (5,4,-1) to the line $\vec{r}=\widehat{i} + \lambda(2\widehat{i} + 9\widehat{j}+5\widehat{k})$

Question 8. Find the foot of the perpendicular drawn from the point $\widehat{i}+6\widehat{j}+3\widehat{k}$ to the line $\vec{r} = (\widehat{j}+2\widehat{k}) + \lambda(\widehat{i}+2\widehat{j}+3\widehat{k})$ . Also, find the length of the perpendicular.

Question 9. Find the equation of the peprendicular drwan from the point P (-1,3,2) to the line $\vec{r} = (2\widehat{j}+3\widehat{k})+\lambda(2\widehat{i}+\widehat{j}+3\widehat{k})$ . Also, find the coordinates of the foot of the perpendicular from P.

Question 10. Find the foot of the perpendicular from (0,2,7) on the line $\frac{x+2}{-1}=\frac{y-1}{3}=\frac{z-3}{-2}$

Question 11. Find the foot of the perpendicular from (1,2,-3) to the line $\frac{x+1}{2}=\frac{y-3}{-2}=\frac{z}{-1}$

Question 13. Find the distance of the point (2,4,-1) from the line $\frac{x+5}{1}=\frac{y+3}{4}=\frac{z-6}{-9}$