1. ⇒ (MHT CET 2023 12th May Evening Shift )

Equation of plane containing the line $\frac{x}{2} = \frac{y}{3} = \frac{z}{4}$ and perpendicular to the plane containing the lines $\frac{x}{3} = \frac{y}{4} = \frac{z}{2}$ and $\frac{x}{4} = \frac{y}{2} = \frac{z}{3}$ is

A. $x + 2 y + z = 0$

B. $x + 2 y - z = 0$

C. $x - 2 y + z = 0$

D. $x - 2 y - z = 0$

Correct answer option is (C)

Equation of the plane containing $\frac{x}{3} = \frac{y}{4} = \frac{z}{2}$ and $\frac{x}{4} = \frac{y}{2} = \frac{z}{3}$ is

$\begin{aligned} | \begin{array}{lll} x & y & z \\ 3 & 4 & 2 \\ 4 & 2 & 3 \end{array} | = 0 \\ \Rightarrow 8 x - y - 10 z = 0 \end{aligned}$

Now required plane is perpendicular to this plane.

Consider option (C)

$(8) (1) + (- 1) (- 2) + (- 10) (1) = 0$

$∴$ Option $(C)$ is correct.

2. ⇒ (MHT CET 2023 12th May Morning Shift )

A plane is parallel to two lines whose direction ratios are $1, 0, - 1$ and $- 1, 1, 0$ and it contains the point $(1, 1, 1)$ . If it cuts the co-ordinate axes at $A, B, C$ , then the volume of the tetrahedron $OABC$ (in cubic units) is

A. $\frac{9}{4}$

B. $\frac{9}{2}$

C. 9

D. 27

Correct answer option is (B)

Equation of the plane passing through $(1, 1, 1)$ is given as

$a (x - 1) + b (y - 1) + c (z - 1) = 0$ .... (i)

As the plane is parallel to the lines having direction ratios $1, 0, - 1$ and $- 1, 1, 0$ , we get $a - c = 0$ and $- a + b = 0$

$\Rightarrow a = b = c$ ..... (ii)

$∴$ From (i) and (ii), we get

$\begin{aligned} x - 1 + y - 1 + z - 1 = 0 \\ ∴ x + y + z = 3 \Rightarrow \frac{x}{3} + \frac{y}{3} + \frac{z}{3} = 1 \end{aligned}$

$∴$ Co-ordinates of A, B, C are $(3, 0, 0), (0, 3, 0)$ and $(0, 0, 3)$ respectively.

$∴$ Volume of tetrahedron $OABC$

$\begin{aligned} = \frac{1}{6} | \begin{array}{lll} 3 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 3 \end{array} | \\ = \frac{1}{6} \times 27 \\ = \frac{9}{2} cu. units \end{aligned}$

3. ⇒ (MHT CET 2023 12th May Morning Shift )

The equation of the plane through $(- 1, 1, 2)$ whose normal makes equal acute angles with co-ordinate axes is

A. $x + y + z - 3 = 0$

B. $x + y + z - 2 = 0$

C. $x + y - z - 2 = 0$

D. $x - y + z - 3 = 0$

Correct answer option is (B)

Note that $(- 1, 1, 2)$ is satisfied by only option (B)

Alternate Method:

Let $A \equiv (- 1, 1, 2)$

$\begin{aligned} ∴ \overset{―}{a} = - \hat{i} + \hat{j} + 2 \hat{k} \\ \overset{―}{n} = \hat{i} + \hat{j} + \hat{k} \end{aligned}$

$∴$ equation of plane is $\overset{―}{r} \cdot \overset{―}{n} = \overset{―}{a} \cdot \overset{―}{n}$

$\begin{aligned} \Rightarrow \bar{r} \cdot (\hat{i} + \hat{j} + \hat{k}) = (- \hat{i} + \hat{j} + 2 \hat{k}) \cdot (\hat{i} + \hat{j} + \hat{k}) \\ \Rightarrow \bar{r} \cdot (\hat{i} + \hat{j} + \hat{k}) = 2 \\ \Rightarrow (x \hat{i} + y \hat{j} + z \hat{k}) \cdot (\hat{i} + \hat{j} + \hat{k}) = 2 \\ \Rightarrow x + y + z - 2 = 0 \end{aligned}$

4. ⇒ (MHT CET 2023 11th May Morning Shift )

The mirror image of the point $(1, 2, 3)$ in a plane is $(- \frac{7}{3}, - \frac{4}{3}, - \frac{1}{3})$ . Thus, the point _________ lies on this plane.

A. $(1, - 1, 1)$

B. $(- 1, - 1, 1)$

C. $(1, 1, 1)$

D. $(- 1, - 1, - 1)$

Correct answer option is (A)

MHT CET 2023 11th May Morning Shift Mathematics - Three Dimensional Geometry Question 18 English Explanation

$M$ is the midpoint.

$∴ M \equiv (- \frac{2}{3}, \frac{1}{3}, \frac{4}{3})$

D.r.s of $A B$ are $\frac{- 10}{3}, \frac{- 10}{3}, \frac{- 10}{3}$ i.e., $1, 1, 1$

Equation of plane is

$\begin{aligned} 1 (x + \frac{2}{3}) + 1 (y - \frac{1}{3}) + 1 (z - \frac{4}{3}) = 0 \\ \Rightarrow x + y + z = 1 \end{aligned}$

Option (A) satisfies this equation of the plane.

5. ⇒ (MHT CET 2023 11th May Morning Shift )

A plane is parallel to two lines, whose direction ratios are $1, 0, - 1$ and $- 1, 1, 0$ and it contains the point $(1, 1, 1)$ . If it cuts co-ordinate axes $(X, Y, Z$ - axes resp.) at $A, B, C$ , then the volume of the tetrahedron $OABC$ is _________ cu. units.

A. 9

B. $\frac{9}{4}$

C. $\frac{9}{2}$

D. 27

Correct answer option is (C)

Equation of the plane passing through $(1, 1, 1)$ is given as

$a (x - 1) + b (y - 1) + c (z - 1) = 0$ .... (i)

As the plane is parallel to the lines having direction ratios $1, 0, - 1$ and $- 1, 1, 0$ , we get

$a - c = 0 and - a + b = 0$

$\Rightarrow a = b = c$ .... (ii)

$∴$ From (i) and (ii), we get

$\begin{aligned} x x - 1 + y - 1 + z - 1 = 0 \\ ∴ x + y + z = 3 \Rightarrow \frac{x}{3} + \frac{y}{3} + \frac{z}{3} = 1 \end{aligned}$

$∴$ Co-ordinates of A, B, C are $(3, 0, 0), (0, 3, 0)$ and $(0, 0, 3)$ respectively.

$\begin{aligned} ∴ Volume of tetrahedron OABC & = \frac{1}{6} | \begin{array}{lll} 3 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 3 \end{array} | \\ = \frac{1}{6} \times 27 \\ = \frac{9}{2} cu. units \end{aligned}$

⇒Line and Plane