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Class 12 RD Sharma Solutions – Chapter 28 The Straight Line in Space – Exercise 28.1 | Set 1

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  • Last Updated : 04 May, 2021
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Question 1. Find the vector and Cartesian equation of the line through the points (5, 2, -4) and which is parallel to the vector3\hat{i}+2\hat{j}-8\hat{k}.

Solution:

As we know that the vector equation of a line is;

\vec{r}=\vec{a}+\lambda\vec{b}

Thus, the Cartesian equation of a line is;

\frac{x-x_1}{a_1}=\frac{y-y_1}{a_2}=\frac{x-x_3}{a_3}

After applying the above formulas;

The vector equation of the line is;

\vec{r}=(5\hat{i}+2\hat{j}-4\hat{k})+\lambda(3\hat{i}+2\hat{j}-8\hat{k})

The Cartesian equation of a line is;

\frac{x-5}{3}=\frac{y-2}{2}=\frac{z+4}{-8}

Question 2. Find the vector equation of the line passing through the points (-1, 0, 2) and (3, 4, 6).

Solution:

Given:

Here, the direction ratios of the line are;

(3 + 1, 4 – 0, 6 – 2) = (4, 4, 4)

Thus, the given line passes through

(-1, 0, 2)

As we know that the vector equation of a line is given as;

\vec{r}=\vec{a}+\lambda\vec{b}

Thus, substitute values

Hence, we get

\vec{r}=\vec{a}+\lambda\vec{b}\\ \vec{r}=(-\vec{i}+0\vec{j}+2\vec{k})+\lambda(4\vec{i}+4\vec{j}+4\vec{k})

Therefore,

Vector equation of the line is;

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

Question 3. Fine the vector equation of a line which is parallel to the vector2\hat{i}-\hat{j}+3\hat{k} and which passes through the point (5, -2, 4), Also, reduce it to Cartesian form.

Solution:

Consider,

The vector equation of line passing through a fixed point vector a and parallel to vector b is shown as;

\vec{r}=\vec{a}+\lambda\vec{b}

Here, λ is scalar

\vec{b}=2\hat{i}-\hat{j}+3\hat{k} and\vec{a}=5\hat{i}-2\hat{j}+4\hat{k}

The equation of the required line is;

\vec{r}=\vec{a}+\lambda\vec{b}\\ \vec{r}=(5\hat{i}-2\hat{j}+4\hat{k})+\lambda(2\hat{i}-\hat{j}+3\hat{k})

Now substitute the value of r here

\vec{r}=x\hat{i}+y\hat{j}+z\hat{k}

Thus, we get

(x\hat{i}+y\hat{j}+z\hat{k})=(5+2\lambda)\hat{i}+(-2-\lambda)\hat{j}+(4+3\lambda)\hat{k}

Now compare the coefficients of vector

x = 5 + 2λ,y = -2 – λ,z = 4 + 3λ

After equating to λ,

We will have

\frac{x-5}{2}=λ ,\ \ \frac{y+2}{-0}=λ ,\ \ \frac{z-4}{3}=λ

Therefore,

The Cartesian form of equation of the line is;

\frac{x-5}{2} =\ \ \frac{y+2}{-0} =\ \ \frac{z-4}{3}

Question 4. A line passing through the point with position vector2\hat{i}-3\hat{j}+4\hat{k} and is in the direction of3\hat{i}+4\hat{j}-5\hat{k} . Find equations of the line in vector and Cartesian form.

Solution:

Consider,

The vector equation of line passing through a fixed point vector a and parallel to vector b is shown as;

\vec{r}=\vec{a}+\lambda\vec{b}

Here, λ is scalar

\vec{a}=2\hat{i}-3\hat{j}+4\hat{k} and\vec{b}=3\hat{i}+4\hat{j}-5\hat{k}

The equation of the required line is;

\vec{r}=\vec{a}+\lambda\vec{b}\\ \vec{r}=(2\hat{i}-3\hat{j}+4\hat{k})+\lambda(3\hat{i}+4\hat{j}-5\hat{k})

Now substitute the value of r here

\vec{r}=x\hat{i}+y\hat{j}+z\hat{k}

Thus, we get

(x\hat{i}+y\hat{j}+z\hat{k})=(2+3\lambda)\hat{i}+(-3+4\lambda)\hat{j}+(4-5\lambda)\hat{k}

Now compare the coefficients of vector

x = 2 + 3λ,y = -3 + 4λ,z = 4 – 5λ

After equating to λ,

We will have

\frac{x-2}{3}=λ ,\ \ \frac{y+3}{4}=λ ,\ \ \frac{z-4}{-5}=λ

Therefore,

The Cartesian form of equation of the line is;

\frac{x-2}{3} =\ \ \frac{y+3}{4} =\ \ \frac{z-4}{-5}

Question 5. ABCD is a parallelogram. The position vectors of the points A, B and C are respectively, 4\hat{i}+5\hat{j}-10\hat{k},\ \ 2\hat{i}-3\hat{j}+4\hat{k} and-\hat{i}+2\hat{j}+\hat{k} . Find the vector equation of the line BD. Also reduce it to Cartesian form.

Solution:

Given: ABCD is a parallelogram.

Consider: AC and BD bisects each other at point O.

Thus,

Position vector of point O =\frac{\vec{a}+\vec{c}}{2}\\ =\frac{(4\hat{i}+5\hat{j}-10\hat{k})+(-\hat{i}+2\hat{j}+\hat{k})}{2}\\ =\frac{3\hat{i}+7\hat{j}-9\hat{k}}{2}

Now, Consider position vector of point O and B are represented by

\vec{o} and\vec{b}

Thus,

Equation of the line BD is the line passing through O and B is given by

\vec{r}=\vec{a}+\lambda(\vec{b}-\vec{a}) [Since equation of the line passing through two points\vec{a} and\vec{b} ]

\vec{r}=\vec{a}+\lambda(\vec{o}-\vec{b})\\ (2\hat{i}-3\hat{j}+4\hat{k})+\lambda\left(\frac{3\hat{i}+7\hat{j}-9\hat{k}}{2}-2\hat{i}-3\hat{j}+4\hat{k}\right)\\ \vec{r}=(2\hat{i}-3\hat{j}+4\hat{k})+\lambda(3\hat{i}+7\hat{j}-9\hat{k}-4\hat{i}+6\hat{j}-8\hat{k})\\ \vec{r}=(2\hat{i}-3\hat{j}+4\hat{k})+\lambda(-\hat{i}+13\hat{j}-17\hat{k})

Now, compare the coefficients of vector i, j, R

x = 2 – λ, y = -3 – 13λ, z = 4 – 17λ

After equating to λ,

We will have

\frac{x-2}{-1}=λ ,\ \ \frac{y+3}{13}=λ ,\ \ \frac{z-4}{-17}=λ

Therefore,

The Cartesian form of equation of the line is;

\frac{x-2}{-1} =\ \ \frac{y+3}{13} =\ \ \frac{z-4}{-17}

Question 6. Find the vector form as well as in Cartesian form, the equation of line passing through the points A(1, 2, -1) and B(2, 1, 1).

Solution:

We know that, equation of line passing though two points (x1, y1 ,z1) and (x2, y2, z2) 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}\ \ \ \ \ \ \ ....(i)\\

Here,

(x1, y1, z1) = A(1, 2, -1)

(x2, y2 ,z2) = B(2, 1, 1)

Using equation (i), equation of line AB,

\frac{x-1}{2-1}=\frac{y-2}{1-2}=\frac{z+1}{1+1}\\ \frac{x-1}{1}=\frac{y-2}{-1}=\frac{z+1}{2}=\lambda\ (assume)

x = λ + 1, y = -λ + 2, z = 2λ – 1

Vector form of equation of line AB is,

x\hat{i}+y\hat{j}+z\hat{k}=(\lambda+1)\hat{i}+(-\lambda+2)\hat{j}+(2\lambda-1)\hat{k}\\ \vec{r}=(\hat{i}+2\hat{j}-\hat{k})+\lambda(\hat{i}-\hat{j}+2\hat{k})

Question 7. Find the vector equation for the line which passes through the point (1, 2, 3) and parallel to the vector\hat{i}-2\hat{j}+3\hat{k}. Reduce the corresponding equation in Cartesian form.

Solution:

We know that vector equation of a line passing through\vec{a} and parallel to the vector\vec{b} is given by,

\vec{r}=\vec{a}+\lambda\vec{b}

Here,

\vec{a}=\hat{i}+2\hat{j}+3\hat{k} and\vec{b}=\hat{i}-2\hat{j}+3\hat{k}

So, required vector equation of line is,

\vec{r}=(\hat{i}+2\hat{j}+3\hat{k})+\lambda(\hat{i}-2\hat{j}+3\hat{k})

Now,

(x\hat{i}+y\hat{j}+z\hat{k})=(1+\lambda)\hat{i}+(2-2\lambda)\hat{j}+(3+3\lambda)\hat{k}

Equating the coefficients of\hat{i},\ \hat{j},\ \hat{k}

x = 1 + λ, y = 2 – 2λ, z = 3 + 3λ

x – 1 = λ,\frac{y-2}{2}=λ,\ \frac{z-3}{3}=λ

So, required equation of line is Cartesian form,

\frac{x-1}{1}=\frac{y-2}{-2}=\frac{z-3}{3}

Question 8. Find the vector equation of a line passing through (2, −1, 1) and parallel to the line whose equations are\frac{x-3}{2}=\frac{y+1}{7}=\frac{z-2}{-3}

Solution:

We know that, equation of a line passing through a point (x1, y1, z1) and having direction ratios proportional to a, b, c is

\frac{x-x_1}{a}=\frac{y-y_1}{b}=\frac{z-z_1}{c}\ \ \ \ .....(i)

Here,

(x1, y1, z1) = (2, -1, 1) and

Given line\frac{x-3}{2}=\frac{y+1}{7}=\frac{z-2}{-3} is parallel to required line.

a = 2μ, b = 7μ, c = -3μ

So, equation of required line using equation (i)

\frac{x-2}{2μ }=\frac{y+1}{7μ }=\frac{z-1}{-3μ }\\ \frac{x-2}{2}=\frac{y+1}{7}=\frac{z-1}{-3}=\lambda

x = 2λ + 2, y = 7λ – 1, z = -3λ + 1

So,

x\hat{i}+y\hat{j}+z\hat{k}=(2λ+2)\hat{i}+(7λ-1)\hat{j}+(-3λ+1)\hat{k}\\ \vec{r}=(2\hat{i}-\hat{j}+\hat{k})+λ(2\hat{i}+7\hat{j}-3\hat{k})

Question 9. The Cartesian equation of a line is\frac{x-5}{3}=\frac{y+4}{7}=\frac{z-6}{2} . Write its vector form

Solution:

The Cartesian equation of the line is

\frac{x-5}{3}=\frac{y+4}{7}=\frac{z-6}{2} ….(i)

The given line passes through the point (5, -4, 6). The position vector of this point is

\vec{a}=5\hat{i}-4\hat{j}+6\hat{k}

Also, the direction ratios of the given line are 3, 7 and 2.

This means that the line is in the direction of vector,

\vec{b}=3\hat{i}+7\hat{j}+2\hat{k}

It is known that the line through position vector\vec{a} and in the direction of the vector\vec{b} is given by the equation,

\vec{r}=\vec{a}+\lambda \vec{b}, \lambda ∈R\\ \vec{r}=(5\hat{i}-4\hat{j}+6\hat{k})+\lambda(3\hat{i}+7\hat{j}+2\hat{k})

Question 10. Find the Cartesian equation of a line passing through (1, -1, 2) and parallel to the line whose equations are \frac{x-3}{1}=\frac{y-1}{2}=\frac{z+1}{-2}. Also, reduce the equation obtained in vector form.

Solution:

We know that, equation of a line passing through a point (x1, y1, z1) and having direction ratios proportional to a, b, c is

\frac{x-x_1}{a}=\frac{y-y_1}{b}=\frac{z-z_1}{c}\ \ \ \ \ ......(i)

Here,

(x1, y1, z1) = (1, -1, 2) and

Given line\frac{x-3}{1}=\frac{y-1}{2}=\frac{z+1}{-2} is parallel to required line,

So,

a = μ, b = 2μ, c = -2μ

So, equation of required line using equation (i) is,

\frac{x-1}{μ }=\frac{y+1}{2μ }=\frac{z-2}{-2μ }\\ \frac{x-1}{1}=\frac{y+1}{2}=\frac{z-2}{-2}=\lambda

x = λ + 1, y = 2λ – 1, z = -2λ +2

So,

x\hat{i}+y\hat{j}+z\hat{k}=(λ+1)\hat{i}+(2λ+1)\hat{j}+(-2λ+ 2)\hat{k}\\ \vec{r}=(\hat{i}-\hat{j}+2\hat{k})+λ(\hat{i}+2\hat{j}-2\hat{k})


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