QBManager

Three Dimensional Geometry

121194 If Cartesian equation of the line is $x - 1 = 2 y +$ $3 = 3 - z$ , then its vector equation is

1

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

2

\vec{r} = (- \hat{i} - 3 \hat{j} + 3 \hat{k}) + λ (\hat{i} + \frac{1}{2} \hat{j} - \hat{k})

3

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

4

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

Explanation:

C Given, Cartesian equation is -
$x - 1 = 2 y + 3 = 3 - z$
$∴ \frac{x - 1}{1} = \frac{2 (y + \frac{3}{2})}{1} = \frac{- (z - 3)}{1} \Rightarrow \frac{x - 1}{1} = \frac{y + \frac{3}{2}}{(\frac{1}{2})} = \frac{z - 3}{- 1}$
$∴ 1, \frac{1}{2}, - 1$ i.e. $2, 1, - 2$ are the direction ratios of the given line.
The given line passes through the point $A (1, \frac{- 3}{2}, 3)$ and parallel to the direction ratios $(1, \frac{1}{2}, - 1)$
$∴$ The vector equation of the given line be
$\vec{r} = \vec{a} + λ \vec{b} (\hat{i} - \frac{3}{2} \hat{j} + 3 \hat{k}) + λ (2 \hat{i} + \hat{j} - 2 \hat{k})$

MHT CET-2020

Three Dimensional Geometry

121195 The equation of the line passing through the point $(1, 2, 3)$ and perpendicular to the lines $\frac{x - 1}{1} = \frac{y - 2}{2} = \frac{z - 3}{3}$ and $\overset{―}{r} = λ (- 3 \hat{i} + 2 \hat{j} + 5 \hat{k})$ is

1

\bar{r} = (\hat{i} + 2 \hat{j} + 3 \hat{k}) + λ (2 \hat{i} - 7 \hat{j} + 4 \hat{k})

2

\bar{r} = (\hat{i} + 2 \hat{j} + 3 \hat{k}) + λ (2 \hat{i} + 7 \hat{j} - 4 \hat{k})

3

\bar{r} = (\hat{i} + 2 \hat{j} + 3 \hat{k}) + λ (2 \hat{i} - 7 \hat{j} - 4 \hat{k})

4

\bar{r} = (\hat{i} + 2 \hat{j} + 3 \hat{k}) + λ (2 \hat{i} + 7 \hat{j} + 4 \hat{k})

Explanation:

A Given, line -
$\frac{x - 1}{1} = \frac{y - 2}{2} = \frac{z - 3}{3} and$
$\bar{r} = λ (- 3 \hat{i} + 2 \hat{j} + 5 \hat{k})$
Let,
${\vec{b}}_{1} = \hat{i} + 2 \hat{j} + 3 \hat{k}$
${\vec{b}}_{2} = - 3 \hat{i} + 2 \hat{j} + 5 \hat{k}$
Now,
$\vec{b} = {\vec{b}}_{1} \times {\vec{b}}_{2}$
$= | \begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & 3 \\ - 3 & 2 & 5 \end{array} |$
$= 4 \hat{i} - 14 \hat{j} + 8 \hat{k} = 2 (2 \hat{i} - 7 \hat{j} + 4 \hat{k})$
Thus, the direction ratios of the required line are $(2, - 7, 4)$ .
The equation of the required line passing through the point $(1, 2, 3)$ is
$\vec{r} = (\hat{i} + 2 \hat{j} + 3 \hat{k}) + λ (2 \hat{i} - 7 \hat{j} + 4 \hat{k}) .$

MHT CET-2020

Three Dimensional Geometry

121198 The equation of the plane passing through the point $(- 1, 2, 1)$ and perpendicular to the line joining the points $(- 3, 1, 2)$ and $(2, 3, 4)$ is

1

\vec{r} (5 \hat{i} - 2 \hat{i} - 2 \hat{k}) = 1

2

\vec{r} (5 \hat{i} + 2 \hat{j} + 2 \hat{k}) = 1

3

\vec{r} (5 \hat{i} + 2 \hat{j} + 2 \hat{k}) = - 1

4

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

Explanation:

B The plane is perpendicular to the line joining the points $B (- 3, 1, 2)$ and $C (2, 3, 4)$ .
Let
$\vec{b} = - 3 \hat{i} + \hat{j} + 2 \hat{k}, \vec{c} = 2 \hat{i} + 3 \hat{j} + 4 \hat{k}$
$\Rightarrow \vec{BC} = \vec{n} = \vec{c} - \vec{b} = 5 \hat{i} + 2 \hat{j} + 2 \hat{k}$
The given plane is passing through the point
$A (- 1, 2, 1) \Rightarrow \vec{a} = - \hat{i} + 2 \hat{j} + \hat{k}$
The equation of the plane is given by $\vec{r} \cdot \vec{n} = \vec{a} \cdot \vec{n}$
$\Rightarrow \vec{r} \cdot (5 \hat{i} + 2 \hat{j} + 2 \hat{k}) = (- \hat{i} + 2 \hat{j} + \hat{k}) \cdot (5 \hat{i} + 2 \hat{j} + 2 \hat{k})$
$\Rightarrow \vec{r} \cdot (5 \hat{i} + 2 \hat{j} + 2 \hat{k}) = - 5 + 4 + 2$ So, equation of the plane is $\vec{r} \cdot (5 \hat{i} + 2 \hat{j} + 2 \hat{k}) = 1$

MHT CET-2019

Three Dimensional Geometry

121199 If planes $\vec{r} \times (p \hat{i} - \hat{j} + 2 \hat{k}) + 3 = 0$ and $\vec{r} \times (\hat{i} - p \hat{j} - \hat{k}) - 5 = 0$ include angle $\frac{π}{3}$ , then the value of $p$ is

1

1, - 3

2

- 1, 3

3 -3

4 3

Explanation:

D We have the planes, $\vec{r} \cdot (p \hat{i} - \hat{j} + 2 \hat{k}) + 3 = 0$ and $\vec{r} \cdot (2 \hat{i} - p - \hat{k}) - 5 = 0$ and the inclination between them is $\frac{π}{3}$ .
Here, ${\vec{n}}_{1} = (p \hat{i} - \hat{j} + 2 \hat{k})$ and ${\vec{n}}_{2} = (2 \hat{i} - p \hat{j} - \hat{k})$
Angle between the planes in their normal form is given by
$\cos θ = | \frac{{\vec{n}}_{1} \cdot {\vec{n}}_{2}}{| {\vec{n}}_{1} | \cdot | n_{2} |} |$
$\Rightarrow \cos \frac{π}{3} = \frac{(p \hat{i} - \hat{j} + 2 \hat{k}) \cdot (2 \hat{i} - p - \hat{j})}{\sqrt{(p)^{2} + (- 1)^{2} + (2)^{2}} \sqrt{(2)^{2} + (- p)^{2} + (- 1)^{2}}}$
$\Rightarrow \frac{1}{2} = \frac{2 p + p - 2}{(\sqrt{p^{2} + 5}) (\sqrt{p^{2} + 5})} \Rightarrow \frac{1}{2} = \frac{3 p - 2}{(p^{2} + 5)}$
$∴ p^{2} + 5 = 6 p - 4 \Rightarrow p^{2} - 6 p + 9 = 0 \Rightarrow (p - 3)^{2} = 0 \Rightarrow p = 3$

MHT CET-2018

Three Dimensional Geometry

121194 If Cartesian equation of the line is $x - 1 = 2 y +$ $3 = 3 - z$ , then its vector equation is

1

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

2

\vec{r} = (- \hat{i} - 3 \hat{j} + 3 \hat{k}) + λ (\hat{i} + \frac{1}{2} \hat{j} - \hat{k})

3

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

4

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

Explanation:

C Given, Cartesian equation is -
$x - 1 = 2 y + 3 = 3 - z$
$∴ \frac{x - 1}{1} = \frac{2 (y + \frac{3}{2})}{1} = \frac{- (z - 3)}{1} \Rightarrow \frac{x - 1}{1} = \frac{y + \frac{3}{2}}{(\frac{1}{2})} = \frac{z - 3}{- 1}$
$∴ 1, \frac{1}{2}, - 1$ i.e. $2, 1, - 2$ are the direction ratios of the given line.
The given line passes through the point $A (1, \frac{- 3}{2}, 3)$ and parallel to the direction ratios $(1, \frac{1}{2}, - 1)$
$∴$ The vector equation of the given line be
$\vec{r} = \vec{a} + λ \vec{b} (\hat{i} - \frac{3}{2} \hat{j} + 3 \hat{k}) + λ (2 \hat{i} + \hat{j} - 2 \hat{k})$

MHT CET-2020

Three Dimensional Geometry

121195 The equation of the line passing through the point $(1, 2, 3)$ and perpendicular to the lines $\frac{x - 1}{1} = \frac{y - 2}{2} = \frac{z - 3}{3}$ and $\overset{―}{r} = λ (- 3 \hat{i} + 2 \hat{j} + 5 \hat{k})$ is

1

\bar{r} = (\hat{i} + 2 \hat{j} + 3 \hat{k}) + λ (2 \hat{i} - 7 \hat{j} + 4 \hat{k})

2

\bar{r} = (\hat{i} + 2 \hat{j} + 3 \hat{k}) + λ (2 \hat{i} + 7 \hat{j} - 4 \hat{k})

3

\bar{r} = (\hat{i} + 2 \hat{j} + 3 \hat{k}) + λ (2 \hat{i} - 7 \hat{j} - 4 \hat{k})

4

\bar{r} = (\hat{i} + 2 \hat{j} + 3 \hat{k}) + λ (2 \hat{i} + 7 \hat{j} + 4 \hat{k})

Explanation:

A Given, line -
$\frac{x - 1}{1} = \frac{y - 2}{2} = \frac{z - 3}{3} and$
$\bar{r} = λ (- 3 \hat{i} + 2 \hat{j} + 5 \hat{k})$
Let,
${\vec{b}}_{1} = \hat{i} + 2 \hat{j} + 3 \hat{k}$
${\vec{b}}_{2} = - 3 \hat{i} + 2 \hat{j} + 5 \hat{k}$
Now,
$\vec{b} = {\vec{b}}_{1} \times {\vec{b}}_{2}$
$= | \begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & 3 \\ - 3 & 2 & 5 \end{array} |$
$= 4 \hat{i} - 14 \hat{j} + 8 \hat{k} = 2 (2 \hat{i} - 7 \hat{j} + 4 \hat{k})$
Thus, the direction ratios of the required line are $(2, - 7, 4)$ .
The equation of the required line passing through the point $(1, 2, 3)$ is
$\vec{r} = (\hat{i} + 2 \hat{j} + 3 \hat{k}) + λ (2 \hat{i} - 7 \hat{j} + 4 \hat{k}) .$

MHT CET-2020

Three Dimensional Geometry

121198 The equation of the plane passing through the point $(- 1, 2, 1)$ and perpendicular to the line joining the points $(- 3, 1, 2)$ and $(2, 3, 4)$ is

1

\vec{r} (5 \hat{i} - 2 \hat{i} - 2 \hat{k}) = 1

2

\vec{r} (5 \hat{i} + 2 \hat{j} + 2 \hat{k}) = 1

3

\vec{r} (5 \hat{i} + 2 \hat{j} + 2 \hat{k}) = - 1

4

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

Explanation:

B The plane is perpendicular to the line joining the points $B (- 3, 1, 2)$ and $C (2, 3, 4)$ .
Let
$\vec{b} = - 3 \hat{i} + \hat{j} + 2 \hat{k}, \vec{c} = 2 \hat{i} + 3 \hat{j} + 4 \hat{k}$
$\Rightarrow \vec{BC} = \vec{n} = \vec{c} - \vec{b} = 5 \hat{i} + 2 \hat{j} + 2 \hat{k}$
The given plane is passing through the point
$A (- 1, 2, 1) \Rightarrow \vec{a} = - \hat{i} + 2 \hat{j} + \hat{k}$
The equation of the plane is given by $\vec{r} \cdot \vec{n} = \vec{a} \cdot \vec{n}$
$\Rightarrow \vec{r} \cdot (5 \hat{i} + 2 \hat{j} + 2 \hat{k}) = (- \hat{i} + 2 \hat{j} + \hat{k}) \cdot (5 \hat{i} + 2 \hat{j} + 2 \hat{k})$
$\Rightarrow \vec{r} \cdot (5 \hat{i} + 2 \hat{j} + 2 \hat{k}) = - 5 + 4 + 2$ So, equation of the plane is $\vec{r} \cdot (5 \hat{i} + 2 \hat{j} + 2 \hat{k}) = 1$

MHT CET-2019

Three Dimensional Geometry

121199 If planes $\vec{r} \times (p \hat{i} - \hat{j} + 2 \hat{k}) + 3 = 0$ and $\vec{r} \times (\hat{i} - p \hat{j} - \hat{k}) - 5 = 0$ include angle $\frac{π}{3}$ , then the value of $p$ is

1

1, - 3

2

- 1, 3

3 -3

4 3

Explanation:

D We have the planes, $\vec{r} \cdot (p \hat{i} - \hat{j} + 2 \hat{k}) + 3 = 0$ and $\vec{r} \cdot (2 \hat{i} - p - \hat{k}) - 5 = 0$ and the inclination between them is $\frac{π}{3}$ .
Here, ${\vec{n}}_{1} = (p \hat{i} - \hat{j} + 2 \hat{k})$ and ${\vec{n}}_{2} = (2 \hat{i} - p \hat{j} - \hat{k})$
Angle between the planes in their normal form is given by
$\cos θ = | \frac{{\vec{n}}_{1} \cdot {\vec{n}}_{2}}{| {\vec{n}}_{1} | \cdot | n_{2} |} |$
$\Rightarrow \cos \frac{π}{3} = \frac{(p \hat{i} - \hat{j} + 2 \hat{k}) \cdot (2 \hat{i} - p - \hat{j})}{\sqrt{(p)^{2} + (- 1)^{2} + (2)^{2}} \sqrt{(2)^{2} + (- p)^{2} + (- 1)^{2}}}$
$\Rightarrow \frac{1}{2} = \frac{2 p + p - 2}{(\sqrt{p^{2} + 5}) (\sqrt{p^{2} + 5})} \Rightarrow \frac{1}{2} = \frac{3 p - 2}{(p^{2} + 5)}$
$∴ p^{2} + 5 = 6 p - 4 \Rightarrow p^{2} - 6 p + 9 = 0 \Rightarrow (p - 3)^{2} = 0 \Rightarrow p = 3$

MHT CET-2018

Three Dimensional Geometry

121194 If Cartesian equation of the line is $x - 1 = 2 y +$ $3 = 3 - z$ , then its vector equation is

1

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

2

\vec{r} = (- \hat{i} - 3 \hat{j} + 3 \hat{k}) + λ (\hat{i} + \frac{1}{2} \hat{j} - \hat{k})

3

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

4

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

Explanation:

C Given, Cartesian equation is -
$x - 1 = 2 y + 3 = 3 - z$
$∴ \frac{x - 1}{1} = \frac{2 (y + \frac{3}{2})}{1} = \frac{- (z - 3)}{1} \Rightarrow \frac{x - 1}{1} = \frac{y + \frac{3}{2}}{(\frac{1}{2})} = \frac{z - 3}{- 1}$
$∴ 1, \frac{1}{2}, - 1$ i.e. $2, 1, - 2$ are the direction ratios of the given line.
The given line passes through the point $A (1, \frac{- 3}{2}, 3)$ and parallel to the direction ratios $(1, \frac{1}{2}, - 1)$
$∴$ The vector equation of the given line be
$\vec{r} = \vec{a} + λ \vec{b} (\hat{i} - \frac{3}{2} \hat{j} + 3 \hat{k}) + λ (2 \hat{i} + \hat{j} - 2 \hat{k})$

MHT CET-2020

Three Dimensional Geometry

121195 The equation of the line passing through the point $(1, 2, 3)$ and perpendicular to the lines $\frac{x - 1}{1} = \frac{y - 2}{2} = \frac{z - 3}{3}$ and $\overset{―}{r} = λ (- 3 \hat{i} + 2 \hat{j} + 5 \hat{k})$ is

1

\bar{r} = (\hat{i} + 2 \hat{j} + 3 \hat{k}) + λ (2 \hat{i} - 7 \hat{j} + 4 \hat{k})

2

\bar{r} = (\hat{i} + 2 \hat{j} + 3 \hat{k}) + λ (2 \hat{i} + 7 \hat{j} - 4 \hat{k})

3

\bar{r} = (\hat{i} + 2 \hat{j} + 3 \hat{k}) + λ (2 \hat{i} - 7 \hat{j} - 4 \hat{k})

4

\bar{r} = (\hat{i} + 2 \hat{j} + 3 \hat{k}) + λ (2 \hat{i} + 7 \hat{j} + 4 \hat{k})

Explanation:

A Given, line -
$\frac{x - 1}{1} = \frac{y - 2}{2} = \frac{z - 3}{3} and$
$\bar{r} = λ (- 3 \hat{i} + 2 \hat{j} + 5 \hat{k})$
Let,
${\vec{b}}_{1} = \hat{i} + 2 \hat{j} + 3 \hat{k}$
${\vec{b}}_{2} = - 3 \hat{i} + 2 \hat{j} + 5 \hat{k}$
Now,
$\vec{b} = {\vec{b}}_{1} \times {\vec{b}}_{2}$
$= | \begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & 3 \\ - 3 & 2 & 5 \end{array} |$
$= 4 \hat{i} - 14 \hat{j} + 8 \hat{k} = 2 (2 \hat{i} - 7 \hat{j} + 4 \hat{k})$
Thus, the direction ratios of the required line are $(2, - 7, 4)$ .
The equation of the required line passing through the point $(1, 2, 3)$ is
$\vec{r} = (\hat{i} + 2 \hat{j} + 3 \hat{k}) + λ (2 \hat{i} - 7 \hat{j} + 4 \hat{k}) .$

MHT CET-2020

Three Dimensional Geometry

121198 The equation of the plane passing through the point $(- 1, 2, 1)$ and perpendicular to the line joining the points $(- 3, 1, 2)$ and $(2, 3, 4)$ is

1

\vec{r} (5 \hat{i} - 2 \hat{i} - 2 \hat{k}) = 1

2

\vec{r} (5 \hat{i} + 2 \hat{j} + 2 \hat{k}) = 1

3

\vec{r} (5 \hat{i} + 2 \hat{j} + 2 \hat{k}) = - 1

4

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

Explanation:

B The plane is perpendicular to the line joining the points $B (- 3, 1, 2)$ and $C (2, 3, 4)$ .
Let
$\vec{b} = - 3 \hat{i} + \hat{j} + 2 \hat{k}, \vec{c} = 2 \hat{i} + 3 \hat{j} + 4 \hat{k}$
$\Rightarrow \vec{BC} = \vec{n} = \vec{c} - \vec{b} = 5 \hat{i} + 2 \hat{j} + 2 \hat{k}$
The given plane is passing through the point
$A (- 1, 2, 1) \Rightarrow \vec{a} = - \hat{i} + 2 \hat{j} + \hat{k}$
The equation of the plane is given by $\vec{r} \cdot \vec{n} = \vec{a} \cdot \vec{n}$
$\Rightarrow \vec{r} \cdot (5 \hat{i} + 2 \hat{j} + 2 \hat{k}) = (- \hat{i} + 2 \hat{j} + \hat{k}) \cdot (5 \hat{i} + 2 \hat{j} + 2 \hat{k})$
$\Rightarrow \vec{r} \cdot (5 \hat{i} + 2 \hat{j} + 2 \hat{k}) = - 5 + 4 + 2$ So, equation of the plane is $\vec{r} \cdot (5 \hat{i} + 2 \hat{j} + 2 \hat{k}) = 1$

MHT CET-2019

Three Dimensional Geometry

121199 If planes $\vec{r} \times (p \hat{i} - \hat{j} + 2 \hat{k}) + 3 = 0$ and $\vec{r} \times (\hat{i} - p \hat{j} - \hat{k}) - 5 = 0$ include angle $\frac{π}{3}$ , then the value of $p$ is

1

1, - 3

2

- 1, 3

3 -3

4 3

Explanation:

D We have the planes, $\vec{r} \cdot (p \hat{i} - \hat{j} + 2 \hat{k}) + 3 = 0$ and $\vec{r} \cdot (2 \hat{i} - p - \hat{k}) - 5 = 0$ and the inclination between them is $\frac{π}{3}$ .
Here, ${\vec{n}}_{1} = (p \hat{i} - \hat{j} + 2 \hat{k})$ and ${\vec{n}}_{2} = (2 \hat{i} - p \hat{j} - \hat{k})$
Angle between the planes in their normal form is given by
$\cos θ = | \frac{{\vec{n}}_{1} \cdot {\vec{n}}_{2}}{| {\vec{n}}_{1} | \cdot | n_{2} |} |$
$\Rightarrow \cos \frac{π}{3} = \frac{(p \hat{i} - \hat{j} + 2 \hat{k}) \cdot (2 \hat{i} - p - \hat{j})}{\sqrt{(p)^{2} + (- 1)^{2} + (2)^{2}} \sqrt{(2)^{2} + (- p)^{2} + (- 1)^{2}}}$
$\Rightarrow \frac{1}{2} = \frac{2 p + p - 2}{(\sqrt{p^{2} + 5}) (\sqrt{p^{2} + 5})} \Rightarrow \frac{1}{2} = \frac{3 p - 2}{(p^{2} + 5)}$
$∴ p^{2} + 5 = 6 p - 4 \Rightarrow p^{2} - 6 p + 9 = 0 \Rightarrow (p - 3)^{2} = 0 \Rightarrow p = 3$

MHT CET-2018

Three Dimensional Geometry

121194 If Cartesian equation of the line is $x - 1 = 2 y +$ $3 = 3 - z$ , then its vector equation is

1

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

2

\vec{r} = (- \hat{i} - 3 \hat{j} + 3 \hat{k}) + λ (\hat{i} + \frac{1}{2} \hat{j} - \hat{k})

3

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

4

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

Explanation:

C Given, Cartesian equation is -
$x - 1 = 2 y + 3 = 3 - z$
$∴ \frac{x - 1}{1} = \frac{2 (y + \frac{3}{2})}{1} = \frac{- (z - 3)}{1} \Rightarrow \frac{x - 1}{1} = \frac{y + \frac{3}{2}}{(\frac{1}{2})} = \frac{z - 3}{- 1}$
$∴ 1, \frac{1}{2}, - 1$ i.e. $2, 1, - 2$ are the direction ratios of the given line.
The given line passes through the point $A (1, \frac{- 3}{2}, 3)$ and parallel to the direction ratios $(1, \frac{1}{2}, - 1)$
$∴$ The vector equation of the given line be
$\vec{r} = \vec{a} + λ \vec{b} (\hat{i} - \frac{3}{2} \hat{j} + 3 \hat{k}) + λ (2 \hat{i} + \hat{j} - 2 \hat{k})$

MHT CET-2020

Three Dimensional Geometry

121195 The equation of the line passing through the point $(1, 2, 3)$ and perpendicular to the lines $\frac{x - 1}{1} = \frac{y - 2}{2} = \frac{z - 3}{3}$ and $\overset{―}{r} = λ (- 3 \hat{i} + 2 \hat{j} + 5 \hat{k})$ is

1

\bar{r} = (\hat{i} + 2 \hat{j} + 3 \hat{k}) + λ (2 \hat{i} - 7 \hat{j} + 4 \hat{k})

2

\bar{r} = (\hat{i} + 2 \hat{j} + 3 \hat{k}) + λ (2 \hat{i} + 7 \hat{j} - 4 \hat{k})

3

\bar{r} = (\hat{i} + 2 \hat{j} + 3 \hat{k}) + λ (2 \hat{i} - 7 \hat{j} - 4 \hat{k})

4

\bar{r} = (\hat{i} + 2 \hat{j} + 3 \hat{k}) + λ (2 \hat{i} + 7 \hat{j} + 4 \hat{k})

Explanation:

A Given, line -
$\frac{x - 1}{1} = \frac{y - 2}{2} = \frac{z - 3}{3} and$
$\bar{r} = λ (- 3 \hat{i} + 2 \hat{j} + 5 \hat{k})$
Let,
${\vec{b}}_{1} = \hat{i} + 2 \hat{j} + 3 \hat{k}$
${\vec{b}}_{2} = - 3 \hat{i} + 2 \hat{j} + 5 \hat{k}$
Now,
$\vec{b} = {\vec{b}}_{1} \times {\vec{b}}_{2}$
$= | \begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & 3 \\ - 3 & 2 & 5 \end{array} |$
$= 4 \hat{i} - 14 \hat{j} + 8 \hat{k} = 2 (2 \hat{i} - 7 \hat{j} + 4 \hat{k})$
Thus, the direction ratios of the required line are $(2, - 7, 4)$ .
The equation of the required line passing through the point $(1, 2, 3)$ is
$\vec{r} = (\hat{i} + 2 \hat{j} + 3 \hat{k}) + λ (2 \hat{i} - 7 \hat{j} + 4 \hat{k}) .$

MHT CET-2020

Three Dimensional Geometry

121198 The equation of the plane passing through the point $(- 1, 2, 1)$ and perpendicular to the line joining the points $(- 3, 1, 2)$ and $(2, 3, 4)$ is

1

\vec{r} (5 \hat{i} - 2 \hat{i} - 2 \hat{k}) = 1

2

\vec{r} (5 \hat{i} + 2 \hat{j} + 2 \hat{k}) = 1

3

\vec{r} (5 \hat{i} + 2 \hat{j} + 2 \hat{k}) = - 1

4

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

Explanation:

B The plane is perpendicular to the line joining the points $B (- 3, 1, 2)$ and $C (2, 3, 4)$ .
Let
$\vec{b} = - 3 \hat{i} + \hat{j} + 2 \hat{k}, \vec{c} = 2 \hat{i} + 3 \hat{j} + 4 \hat{k}$
$\Rightarrow \vec{BC} = \vec{n} = \vec{c} - \vec{b} = 5 \hat{i} + 2 \hat{j} + 2 \hat{k}$
The given plane is passing through the point
$A (- 1, 2, 1) \Rightarrow \vec{a} = - \hat{i} + 2 \hat{j} + \hat{k}$
The equation of the plane is given by $\vec{r} \cdot \vec{n} = \vec{a} \cdot \vec{n}$
$\Rightarrow \vec{r} \cdot (5 \hat{i} + 2 \hat{j} + 2 \hat{k}) = (- \hat{i} + 2 \hat{j} + \hat{k}) \cdot (5 \hat{i} + 2 \hat{j} + 2 \hat{k})$
$\Rightarrow \vec{r} \cdot (5 \hat{i} + 2 \hat{j} + 2 \hat{k}) = - 5 + 4 + 2$ So, equation of the plane is $\vec{r} \cdot (5 \hat{i} + 2 \hat{j} + 2 \hat{k}) = 1$

MHT CET-2019

Three Dimensional Geometry

121199 If planes $\vec{r} \times (p \hat{i} - \hat{j} + 2 \hat{k}) + 3 = 0$ and $\vec{r} \times (\hat{i} - p \hat{j} - \hat{k}) - 5 = 0$ include angle $\frac{π}{3}$ , then the value of $p$ is

1

1, - 3

2

- 1, 3

3 -3

4 3

Explanation:

D We have the planes, $\vec{r} \cdot (p \hat{i} - \hat{j} + 2 \hat{k}) + 3 = 0$ and $\vec{r} \cdot (2 \hat{i} - p - \hat{k}) - 5 = 0$ and the inclination between them is $\frac{π}{3}$ .
Here, ${\vec{n}}_{1} = (p \hat{i} - \hat{j} + 2 \hat{k})$ and ${\vec{n}}_{2} = (2 \hat{i} - p \hat{j} - \hat{k})$
Angle between the planes in their normal form is given by
$\cos θ = | \frac{{\vec{n}}_{1} \cdot {\vec{n}}_{2}}{| {\vec{n}}_{1} | \cdot | n_{2} |} |$
$\Rightarrow \cos \frac{π}{3} = \frac{(p \hat{i} - \hat{j} + 2 \hat{k}) \cdot (2 \hat{i} - p - \hat{j})}{\sqrt{(p)^{2} + (- 1)^{2} + (2)^{2}} \sqrt{(2)^{2} + (- p)^{2} + (- 1)^{2}}}$
$\Rightarrow \frac{1}{2} = \frac{2 p + p - 2}{(\sqrt{p^{2} + 5}) (\sqrt{p^{2} + 5})} \Rightarrow \frac{1}{2} = \frac{3 p - 2}{(p^{2} + 5)}$
$∴ p^{2} + 5 = 6 p - 4 \Rightarrow p^{2} - 6 p + 9 = 0 \Rightarrow (p - 3)^{2} = 0 \Rightarrow p = 3$

MHT CET-2018