RD Sharma Class 12 Ex 28.1 Solutions Chapter 28 The Straight Line in Space

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Textbook	NCERT
Class	Class 12th
Subject	Maths
Chapter	28
Exercise	28.1
Category	RD Sharma Solutions

Table of Contents

RD Sharma Class 12 Ex 28.1 Solutions Chapter 28 The Straight Line in Space

Question 1: Ten cards numbered 1 through 10 are placed in a box, mixed

Question 1. Find the vector and Cartesian equation of the line through the points (5, 2, -4) and which is parallel to the vector $3\hat{i}+2\hat{j}-8\hat{k}.$

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 vector $2\hat{i}-\hat{j}+3\hat{k}$ 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 vector $2\hat{i}-3\hat{j}+4\hat{k}$ and is in the direction of $3\hat{i}+4\hat{j}-5\hat{k}$ . 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,

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

Question 11. Find the direction cosines of the line $\frac{4-x}{2}=\frac{y}{6}=\frac{1-z}{3}$ . Also, reduce it to vector form

Solution:

Given:

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

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

x = -2λ + 4, y = 6λ, z = -3λ + 1

So,

$x\hat{i}+y\hat{j}+z\hat{k}=(-2λ+4)\hat{i}+(6λ)\hat{j}+(-3λ+1)\hat{k}\\ \vec{r}=(4\hat{i}+\hat{k})+λ(-2\hat{i}+6\hat{j}-3\hat{k})$

Direction ratios of the line are = -2, 6, -3

Direction cosines of the lines are,

$\frac{a}{\sqrt{a^2+b^2+c^2}},\ \frac{b}{\sqrt{a^2+b^2+c^2}},\ \frac{c}{\sqrt{a^2+b^2+c^2}}\\ \frac{-2}{\sqrt{(-2)^2+(6)^2+(-3)^2}},\ \frac{6}{\sqrt{(-2)^2+6^2+(-3)^2}},\ \frac{-3}{\sqrt{(-2)^2+6^2+(-3)^2}}\\ \frac{-2}{7},\ \frac{6}{7},\ \frac{-3}{7}$

Question 12. The Cartesian equations of a line are x = ay + b, z = cy + d. Find its direction ratios and reduce it to vector form.

Solution:

x = ay + b

z = cy + d

$\frac{x-b}{a}=\frac{y}{1}=\frac{z-d}{c}=\lambda$

So, DR’s of line are (a, 1, c)

From above equation, we can write

x = aλ + b

y = λ

z = cλ + d

So vector equation of line is

$x\hat{i}+y\hat{j}+z\hat{k}=(b\hat{i}+d\hat{k})+λ(a\hat{i}+\hat{j}+c\hat{k})$

Question 13. Find the vector equation of a line passing through the point with position vector $\hat{i}-2\hat{j}-3\hat{k}$ and parallel to the line joining the points with the position vector $\hat{i}-\hat{j}+4\hat{k}$ and $2\hat{i}+\hat{j}+2\hat{k}$ . Also, find the Cartesian equivalent of this equation.

Solution:

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

$\vec{r}=\vec{a}+λ\vec{b}$ ……. (i)

Here,

$\vec{a}=\hat{i}-2\hat{j}-3\hat{k}$

and, \vec{b} = line joining $(\hat{i}-\hat{j}+4\hat{k})$ and $(2\hat{i}+\hat{j}+2\hat{k})$

$=(2\hat{i}+\hat{j}+2\hat{k})-(\hat{i}-\hat{j}+4\hat{k})\\ =2\hat{i}-\hat{i}+\hat{j}+\hat{j}+2\hat{k}-4\hat{k}\\ =\hat{i}+2\hat{j}-2\hat{k}$

Equation of the line is

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

For Cartesian form of equation put $x\hat{i}+y\hat{j}+z\hat{k}$

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

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

x = 1 + λ, y = -2 + 2λ, z = -3 – 2λ

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

Question 14. Find the points on the line $\frac{x+2}{3}=\frac{y+1}{2}=\frac{z-3}{2}$ at a distance of 5 units from the points P(1, 3, 3).

Solution:

Given, line is $\frac{x+2}{3}=\frac{y+1}{2}=\frac{z-3}{2}=\lambda$
General points Q on line is (3λ – 2, 2λ -1), 2λ + 3)
Distance of points P from Q = $\sqrt{(x_1-x_2)^2+(y_1-y_2)^2+(z_1-z_2)^2}$
PQ = $\sqrt{(3λ-2-1)^2+(2λ-1-3)^2+(2λ+3-3)^2}$
(5)² = (3λ -3)² + (2λ – 4) + (2λ)²
25 = 9λ² + 9 – 18λ + 4λ² + 16 – 16λ + 4λ²
17λ² – 34λ = 0
17λ (λ – 2) = 0
λ = 0 or 2
So, points on the line are (3(0) – 2, 2(0) – 1, 2(0) + 3)
(3(2) – 2, 2(2) – 1, 2(2) + 3)
= (-2, -1, 3), (4, 3, 7)

Question 15. Show that the points whose position vectors are $-2\hat{i}+3\hat{j},\ \hat{i} + 2\hat{j}+3\hat{k}$ and $7\hat{i}-\hat{k}$ are collinear.

Solution:

Let the given points are A,B.C with position vectors $\vec{a},\ \vec{b},\ \vec{c}$ respectively.

$\vec{a}=2\hat{i}+3\hat{j},\ \vec{b}=\hat{i}+2\hat{j}+3\hat{k},\ \vec{c}=7\hat{i}-\hat{k}$

We know that, equation of a line passing through $\vec{a}$ and $\vec{b}$ are

$\vec{r}=\vec{a}+\lambda(\vec{b}-\vec{a})\\ =(-2\hat{i}+3\hat{j})+λ((\hat{i}+2\hat{j}+3\hat{k})-(-2\hat{i}+3\hat{j}))\\ =(-2\hat{i}+3\hat{j})+\lambda(\hat{i}+2\hat{j}+3\hat{k}+2\hat{i}-3\hat{j})\\ \vec{r}=(-2\hat{i}+3\hat{j})+\lambda(3\hat{i}-\hat{j}+3\hat{k})\ \ \ \ .....(i)$

If A, B, C are collinear then $\vec{c}$ must satisfy equation (i)

$7\hat{i}-\hat{k}=(-2+3\lambda)\hat{i}+(3-\lambda)\hat{j}+(3\lambda)\hat{k}$

Equation the coefficients of $\vec{i},\ \vec{j},\ \vec{k}$

-2 + 3 = 7 , λ = 3

3 – λ = 0 , λ = 3

3λ = -1 , λ = $-\frac{1}{3}$

Since, value of λ are not equal, so,

Given points are collinear.

Question 16. Find the Cartesian and vector equations of a line which passes through the points (1, 2, 3) and is parallel to the line $\frac{-x-2}{1}=\frac{y+3}{7}=\frac{2z-6}{3}$

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) = (1, 2, 3) and

Given line $\frac{-x-2}{1}=\frac{y+3}{7}=\frac{2z-6}{3}$

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

Its parallel to the required line, so

a = μ , b = 7μ, c = $\frac{3}{2}$ μ

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

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

Multiplying the denominators by 2

$\frac{x-1}{-2}=\frac{y-2}{14}=\frac{z-3}{3}=λ$

x = -2λ + 1, y = 14λ + 2, z = 3λ + 3

So, vector form of the equation of required line,

$x\hat{i}+y\hat{j}+z\hat{k}=(-2λ+1)\hat{i}+(14λ+2)\hat{j}+(3λ+3)\hat{k}\\ \vec{r}=(\hat{i}+2\hat{j}+3\hat{k})+λ(-2\hat{i}+14\hat{j}+3\hat{k})$

Question 17. The Cartesian equations of a line are 3x + 1 = 6y – 2 = 1 – z. Find the fixed point through which it passes, its direction ratios, and also its vector equation.

Solution:

Given equation of line is,

3x + 1 = 6y -2 = 1 – z

Dividing all by 6

$=\frac{3x+1}{6}=\frac{6y-2}{6}=\frac{1-z}{6}\\ =\frac{3x}{6}+\frac{1}{6}=\frac{6y}{6}-\frac{2}{6}=\frac{1}{6}-\frac{z}{6}\\ =\frac{1}{2}x+\frac{1}{6}=y-\frac{1}{3}=-\frac{z}{6}+\frac{1}{6}\\ =\frac{1}{2}\left(x+\frac{1}{3}\right)=1\left(y-\frac{1}{3}\right)=\frac{1}{6}(z-1)\\ =\frac{x+\frac{1}{3}}{2}=\frac{y-\frac{1}{3}}{1}=\frac{z-1}{-6}=λ$

Comparing it with equation of line equation of line passing through (x,₁ y₁, z₁) and the direction ratios a, b, c,

$\frac{x-x_1}{a}=\frac{y-y_1}{b}=\frac{z-z_1}{c}\\ (x_1,\ y_1,\ z_1)=\left(-\frac{1}{3},\ \frac{1}{3},\ 1\right)$

a = 2, b = 1, -6

So, direction ratios of the line are -2, 1, -6

From equation (i)

$x = \left(2λ-\frac{1}{3}\right),\ y=\left(λ+\frac{1}{3}\right),\ z=(-6λ+1)$

So, vector equation of the given line is,

$x\hat{i}+y\hat{j}+z\hat{k}=\left(2λ-\frac{1}{3}\right)\hat{i}+\left(λ+\frac{1}{3}\right)\hat{j}+(-6λ+1)\hat{k}\\ \vec{r}=\left(-\frac{1}{3}\hat{i}+\frac{1}{3}\hat{j}+\hat{k}\right)+λ(2\hat{i}+\hat{j}-6\hat{k})$

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RD Sharma Class 12 Ex 28.1 Solutions Chapter 28 The Straight Line in Space

Question 1: Ten cards numbered 1 through 10 are placed in a box, mixed

Question 1. Find the vector and Cartesian equation of the line through the points (5, 2, -4) and which is parallel to the vector

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

Question 3. Fine the vector equation of a line which is parallel to the vectorand which passes through the point (5, -2, 4), Also, reduce it to Cartesian form.

Question 4. A line passing through the point with position vectorand is in the direction of. Find equations of the line in vector and Cartesian form.

Question 5. ABCD is a parallelogram. The position vectors of the points A, B and C are respectively, and. Find the vector equation of the line BD. Also reduce it to Cartesian form.

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).

Question 7. Find the vector equation for the line which passes through the point (1, 2, 3) and parallel to the vector. Reduce the corresponding equation in Cartesian form.

Question 8. Find the vector equation of a line passing through (2, −1, 1) and parallel to the line whose equations are

Question 9. The Cartesian equation of a line is. Write its vector form

Question 10. Find the Cartesian equation of a line passing through (1, -1, 2) and parallel to the line whose equations are . Also, reduce the equation obtained in vector form.

Question 11. Find the direction cosines of the line. Also, reduce it to vector form

Question 12. The Cartesian equations of a line are x = ay + b, z = cy + d. Find its direction ratios and reduce it to vector form.

Question 13. Find the vector equation of a line passing through the point with position vectorand parallel to the line joining the points with the position vectorand. Also, find the Cartesian equivalent of this equation.

Question 14. Find the points on the lineat a distance of 5 units from the points P(1, 3, 3).

Question 15. Show that the points whose position vectors areandare collinear.

Question 16. Find the Cartesian and vector equations of a line which passes through the points (1, 2, 3) and is parallel to the line

Question 17. The Cartesian equations of a line are 3x + 1 = 6y – 2 = 1 – z. Find the fixed point through which it passes, its direction ratios, and also its vector equation.

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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 vector $3\hat{i}+2\hat{j}-8\hat{k}.$

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

Question 4. A line passing through the point with position vector $2\hat{i}-3\hat{j}+4\hat{k}$ and is in the direction of $3\hat{i}+4\hat{j}-5\hat{k}$ . Find equations of the line in vector and Cartesian form.

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.

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.

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

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

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.

Question 11. Find the direction cosines of the line $\frac{4-x}{2}=\frac{y}{6}=\frac{1-z}{3}$ . Also, reduce it to vector form

Question 14. Find the points on the line $\frac{x+2}{3}=\frac{y+1}{2}=\frac{z-3}{2}$ at a distance of 5 units from the points P(1, 3, 3).

Question 15. Show that the points whose position vectors are $-2\hat{i}+3\hat{j},\ \hat{i} + 2\hat{j}+3\hat{k}$ and $7\hat{i}-\hat{k}$ are collinear.

Question 16. Find the Cartesian and vector equations of a line which passes through the points (1, 2, 3) and is parallel to the line $\frac{-x-2}{1}=\frac{y+3}{7}=\frac{2z-6}{3}$