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Class 12 RD Sharma Solutions- Chapter 28 The Straight Line in Space – Exercise 28.4
  • Difficulty Level : Medium
  • Last Updated : 19 Jan, 2021

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 perepndicular 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 thr 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 vecor 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 th egiven 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)

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