JEE Mathematics 2024 Question with Solution

Given: The point is $P(a,a,a)$. Point $Q$ is the foot of the perpendicular from $P$ to the line $x=y, z=1$, and point $R$ is the foot of the perpendicular from $P$ to the line $x=-y, z=-1$.

Find: The value of $12a^2$ when $\angle QPR = 90^\circ$.

From the line $x=y, z=1$, write

$$Q(r,r,1)$$

From the line $x=-y, z=-1$, write

$$R(k,-k,-1)$$

the solution gives

$$\overrightarrow{PQ}=(a-r)\hat{i}+(a-r)\hat{j}+(a-1)\hat{k}$$

and

$$\overrightarrow{PR}=(a-k)\hat{i}+(a+k)\hat{j}+(a+1)\hat{k}$$

Since $\overrightarrow{PQ} \perp \overrightarrow{PR}$,

$$(a-r)(a-k)+(a-r)(a+k)+(a-1)(a+1)=0$$

On simplifying, the solution concludes

$$a=1 \text{ or } -1$$

Therefore,

$$12a^2=12$$

Hence, the correct option is A.

Given: $P(a,a,a)$ and the lines $x=y, z=1$ and $x=-y, z=-1$.

Find: $12a^2$ using the right-angle condition at $P$.

For the first line, use parametric form

$$(x,y,z)=(t,t,1)$$

So the foot of the perpendicular is

$$Q=(a,a,1)$$

For the second line, use parametric form

$$(x,y,z)=(s,-s,-1)$$

The perpendicular condition gives

$$(s-a,-s-a,-1-a)\cdot(1,-1,0)=0$$

which yields

$$s=0$$

Hence,

$$R=(0,0,-1)$$

Now,

$$\overrightarrow{PQ}=Q-P=(0,0,1-a)$$

and

$$\overrightarrow{PR}=R-P=(-a,-a,-1-a)$$

Since $\angle QPR=90^\circ$,

$$\overrightarrow{PQ}\cdot\overrightarrow{PR}=0$$

So,

$$0\cdot(-a)+0\cdot(-a)+(1-a)(-1-a)=0$$

Thus,

$$(1-a)(-1-a)=0$$

which gives

$$a=-1$$

Therefore,

$$12a^2=12(-1)^2=12$$

So the correct option is A.