JEE Mathematics 2023 Question with Solution

Given: The plane passes through $(0,-1,2)$ and $(-1,2,1)$, and is parallel to the line through $(5,1,-7)$ and $(1,-1,-1)$.

Find: Which given point also lies on the plane.

First, form two vectors lying in the plane:

$$\mathbf{a} = (-1-0,\; 2-(-1),\; 1-2) = (-1,3,-1)$$ $$\mathbf{b} = (1-5,\; -1-1,\; -1-(-7)) = (-4,-2,6)$$

Since the plane contains $\mathbf{a}$ and is parallel to $\mathbf{b}$, its normal vector is

$$\mathbf{n} = \mathbf{a} \times \mathbf{b}$$

Using the cross product,

$$\mathbf{n} = (16,10,14)$$

Now use the point-normal form of the plane through $(0,-1,2)$:

$$16(x-0) + 10(y+1) + 14(z-2) = 0$$

Simplifying,

$$16x + 10y + 14z - 18 = 0$$

Check option $B$, namely $(-2,5,0)$:

$$16(-2) + 10(5) + 14(0) - 18 = -32 + 50 - 18 = 0$$

So this point satisfies the plane equation.

Therefore, the correct option is B, that is, $(-2,5,0)$.

Given: Two points on the plane are $(0,-1,2)$ and $(-1,2,1)$, and the plane is parallel to the line through $(5,1,-7)$ and $(1,-1,-1)$.

Find: The option that lies on this plane.

A vector joining the two given points of the plane is

$$\mathbf{a} = (-1,3,-1)$$

A direction vector of the given line is

$$\mathbf{b} = (-4,-2,6)$$

The normal vector is obtained by

$$\mathbf{n} = \mathbf{a} \times \mathbf{b}$$

Expand it as

$$\begin{aligned} \mathbf{n} &= \hat{i}\big((3)(6)-(-1)(-2)\big) - \hat{j}\big((-1)(6)-(-1)(-4)\big) + \hat{k}\big((-1)(-2)-(3)(-4)\big) \\ &= \hat{i}(18-2) - \hat{j}(-6-4) + \hat{k}(2+12) \\ &= 16\hat{i} + 10\hat{j} + 14\hat{k} \end{aligned}$$

Hence,

$$\mathbf{n} = (16,10,14)$$

Equation of the plane through $(0,-1,2)$ is

$$16(x-0) + 10(y+1) + 14(z-2) = 0$$ $$16x + 10y + 14z - 18 = 0$$

Now test the options. For $(-2,5,0)$,

$$16(-2) + 10(5) + 14(0) - 18 = 0$$

Hence this point lies on the plane.

Therefore, the correct option is B.