JEE Mathematics 2023 Question with Solution

Given: The system of equations is

$$\begin{aligned} -x+2y-9z &= 7 \\ -x+3y+7z &= 9 \\ -2x+y+5z &= 8 \\ -3x+y+13z &= \lambda \end{aligned}$$

Find: The distance of the point $(\alpha,\beta,\gamma)$ from the plane $2x-2y+z=\lambda$.

First solve the first three equations. From

$$-x+2y-9z=7 \quad (1)$$ $$-x+3y+7z=9 \quad (2)$$ $$-2x+y+5z=8 \quad (3)$$

Subtracting $(1)$ from $(2)$,

$$y+16z=2 \quad (4)$$

Multiply $(1)$ by $2$ and subtract $(3)$:

$$(-2x+4y-18z)-(-2x+y+5z)=14-8$$

so

$$3y-23z=6 \quad (5)$$

Now solve $(4)$ and $(5)$. From $(4)$,

$$y=2-16z$$

Substitute into $(5)$:

$$3(2-16z)-23z=6$$ $$6-48z-23z=6$$ $$71z=0 \Rightarrow z=0$$

Hence,

$$y=2$$

Substitute in $(1)$:

$$-x+4=7 \Rightarrow x=-3$$

Thus,

$$(\alpha,\beta,\gamma)=(-3,2,0)$$

Now substitute this solution into the fourth equation to find $\lambda$:

$$-3(-3)+2+13(0)=\lambda$$ $$9+2=\lambda \Rightarrow \lambda=11$$

Use the distance formula from the point $(x_1,y_1,z_1)$ to the plane $Ax+By+Cz+D=0$:

$$d=\frac{|Ax_1+By_1+Cz_1+D|}{\sqrt{A^2+B^2+C^2}}$$

The plane is

$$2x-2y+z-\lambda=0$$

Substitute $(x_1,y_1,z_1)=(-3,2,0)$ and $\lambda=11$:

$$d=\frac{|2(-3)-2(2)+0-11|}{\sqrt{4+4+1}}$$ $$=\frac{|-6-4-11|}{3}$$ $$=\frac{21}{3}=7$$

Therefore, the distance is $7$ and the correct option is A.

Given: There are more equations than variables. Find: The required distance.

A quick way is to first solve any consistent subset of three equations to get the unique point, then use the remaining equation to determine the parameter.

From the first three equations, the solution is

$$(\alpha,\beta,\gamma)=(-3,2,0)$$

Using the fourth equation,

$$-3(-3)+2+13(0)=\lambda$$ $$\lambda=11$$

Now the plane becomes

$$2x-2y+z-11=0$$

Distance from $(-3,2,0)$ is

$$d=\frac{|2(-3)-2(2)+0-11|}{\sqrt{2^2+(-2)^2+1^2}}=\frac{21}{3}=7$$

Therefore, the correct option is A.