6. ⇒ (MHT CET 2023 10th May Morning Shift )

The co-ordinates of the point, where the line through $A (3, 4, 1)$ and $B (5, 1, 6)$ crosses the $XZ$ -plane, are

A. $(\frac{11}{3}, 0, \frac{21}{3})$

B. $(\frac{17}{3}, 0, \frac{23}{3})$

C. $(\frac{- 11}{3}, 0, \frac{21}{3})$

D. $(\frac{17}{3}, 0, \frac{- 23}{3})$

Correct answer option is (B)

Let $A (x_{1}, y_{1}, z_{1}) = A (3, 4, 1)$ and $B (x_{2}, y_{2}, z_{2}) = B (5, 1, 6)$

The equation of the line passing through the points $(x_{1}, y_{1}, z_{1})$ and $(x_{2}, y_{2}, z_{2})$ is given by

$\frac{x - x_{1}}{x_{2} - x_{1}} = \frac{y - y_{1}}{y_{2} - y_{1}} = \frac{z - z_{1}}{z_{2} - z_{1}}$

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

$∴ \frac{x - 3}{2} = \frac{y - 4}{- 3} = \frac{z - 1}{5}$

Since the line crosses the $XZ$ plane, $y = 0$

$\begin{array}{ll} ∴ & \frac{x - 3}{2} = \frac{4}{3} = \frac{z - 1}{5} \\ ∴ & \frac{x - 3}{2} = \frac{4}{3} and \frac{z - 1}{5} = \frac{4}{3} \\ \Rightarrow x = \frac{17}{3} and z = \frac{23}{3} \end{array}$

$∴$ The required point is $(\frac{17}{3}, 0, \frac{23}{3})$ .

7. ⇒ (MHT CET 2023 9th May Morning Shift )

If the Cartesian equation of a line is $6 x - 2 = 3 y + 1 = 2 z - 2$ , then the vector equation of the line is

A. $\overset{―}{r} = (\frac{1}{3} \hat{i} - \frac{1}{3} \hat{j} + \hat{k}) + λ (\hat{i} + 2 \hat{j} + 3 \hat{k})$

B. $\overset{―}{r} = (\hat{i} + \hat{j} + \hat{k}) + λ (\hat{i} + 2 \hat{j} + 3 \hat{k})$

C. $\overset{―}{r} = (\frac{- 1}{3} \hat{i} + \frac{1}{3} \hat{j} + \hat{k}) + λ (\hat{i} - 2 \hat{j} + 3 \hat{k})$

D. $\overset{―}{r} = (\frac{1}{3} \hat{i} - \frac{1}{3} \hat{j} - \hat{k}) + λ (\hat{i} - \hat{j} + \hat{k})$

Correct answer option is (A)

Given Cartesian equation of the line is

$\begin{aligned} 6 x - 2 = 3 y + 1 = 2 z - 2 \\ \Rightarrow 6 (x - \frac{1}{3}) = 3 (y + \frac{1}{3}) = 2 (z - 1) \end{aligned}$

$\begin{aligned} \Rightarrow & \frac{x - \frac{1}{3}}{\frac{1}{6}} = \frac{y + \frac{1}{3}}{\frac{1}{3}} = \frac{z - 1}{\frac{1}{2}} \\ \Rightarrow & \frac{x - \frac{1}{3}}{1} = \frac{y + \frac{1}{3}}{2} = \frac{z - 1}{3} \end{aligned}$

$∴$ The given line passes through $(\frac{1}{3}, \frac{- 1}{3}, 1)$ and has direction ratios proportional to $1, 2, 3$ .

$∴$ Vector equation is

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

8. ⇒ (MHT CET 2021 21th September Evening Shift )

The direction cosines $ℓ, m, n$ of the line $\frac{x + 2}{2} = \frac{2 y - 5}{3}; z = - 1$ are

A. $ℓ = \pm \frac{1}{\sqrt{5}}, m = 0, n = \pm \frac{2}{\sqrt{5}}$

B. $ℓ = \pm \frac{3}{5}, m = \pm \frac{4}{5}, n = 0$

C. $ℓ = \pm \frac{4}{5}, m = \pm \frac{3}{5}, n = 0$

D. $ℓ = \pm \frac{1}{\sqrt{3}}, m = \pm \frac{1}{\sqrt{3}}, n = \pm \frac{1}{\sqrt{3}}$

Correct answer option is (C)

We have line $\frac{x + 2}{2} = \frac{2 y - 5}{3}, z = - 1$

i.e. $\frac{x - (- 2)}{2} = \frac{2 (y - \frac{5}{2})}{3}, z = - 1$

i.e. $\frac{x - (- 2)}{2} = \frac{(y - \frac{5}{2})}{(\frac{3}{2})}, z = - 1$

Here direction ratios are $2, \frac{3}{2}, 0$

Also $\sqrt{(2)^{2} + {(\frac{3}{2})}^{2} + 0} = \pm \frac{5}{2}$

Here required direction cosines are

$\frac{2}{(\pm \frac{5}{2})}, \frac{(\frac{3}{2})}{(\pm \frac{5}{2})}, \frac{0}{(\pm \frac{5}{2})} i.e. \pm \frac{4}{5}, \pm \frac{3}{5}, 0$

9. ⇒ (MHT CET 2021 21th September Evening Shift )

The equation of a line passing through $(3, - 1, 2)$ and perpendicular to the lines $\bar{r} = (\hat{i} + \hat{j} - \hat{k}) + λ (2 \hat{i} - 2 \hat{j} + \hat{k})$ and $\bar{r} = (2 \hat{i} + \hat{j} - 3 \hat{k}) + μ (\hat{i} - 2 \hat{j} + 2 \hat{k})$ is

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

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

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

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

Correct answer option is (A)

Let $a, b, c$ be the direction ratios of the required line.

$∴ 2 a - 2 b + c = 0$ ....... (1) and $a - 2 b + 2 c = 0$ ..... (2)

From (1) and (2), we write

$\begin{aligned} \frac{a}{| \begin{array}{ll} - 2 & 1 \\ - 2 & 2 \end{array} |} = \frac{- b}{| \begin{array}{ll} 2 & 1 \\ 1 & 2 \end{array} |} = \frac{c}{| \begin{array}{ll} 2 & - 2 \\ 1 & - 2 \end{array} |} \Rightarrow \frac{a}{- 2} = \frac{b}{- 3} = \frac{c}{- 2} \\ ∴ (a, b, c) = (2, 3, 2) \end{aligned}$

So equation of required line is

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

10. ⇒ (MHT CET 2021 20th September Evening Shift )

The Cartesian equation of the line passing through the points A(2, 2, 1) and B(1, 3, 0) is

A. $\frac{x + 2}{1} = \frac{y + 2}{- 1} = \frac{z + 1}{- 1}$

B. $\frac{x - 2}{- 1} = \frac{y - 2}{1} = \frac{z - 1}{- 1}$

C. $\frac{x + 2}{- 1} = \frac{y + 2}{1} = \frac{z + 1}{- 1}$

D. None of these

Correct answer option is (B)

The required Cartesian equation of line is

$\frac{x - 2}{1 - 2} = \frac{y - 2}{3 - 2} = \frac{z - 1}{0 - 1} i.e. \frac{x - 2}{- 1} = \frac{y - 2}{1} = \frac{z - 1}{- 1}$

⇒Line and Plane