Let P be the image of the point Q(7,-2,5) in the line L: \fracx-12=\fracy+13=\fracz4 and R(5, p, q) be a point on L . Then the square of the area of \triangle P Q R is \underline\hspace2cm .

3D Geometry PYQ — JEE Main 2025 · 24 Jan · Shift 2 · Q46 with Solution

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Correct answer: 957

Interpretation of “image of the point in the line”

The image of point $Q$ in the line $L$ means the reflection of $Q$ about the line $L$ .

So if $M$ is the foot of the perpendicular from $Q$ to the line $L$ , then $M$ is the midpoint of $PQ$ and $PM = QM, \quad P = 2M - Q.$

Parametric form of the line

Given $\frac{x-1}{2}=\frac{y+1}{3}=\frac{z}{4}=t,$ so a general point on $L$ is

-1+3t, 4t).$$ The line passes through $$A_0=(1,-1,0)$$ and has direction vector $$\vec d=(2,3,4).$$ Point $$Q=(7,-2,5).$$ --- 3. **Find the foot of perpendicular $M$ from $Q$ to $L$** Let $$M=(1+2t,-1+3t,4t).$$ Then $$\overrightarrow{QM}=M-Q=(1+2t-7,-1+3t+2,4t-5)=(2t-6,3t+1,4t-5).$$ Since $QM \perp L$, we must have $$\overrightarrow{QM}\cdot \vec d=0.$$ So, $$(2t-6)\cdot 2+(3t+1)\cdot 3+(4t-5)\cdot 4=0.$$ Compute: $$4t-12+9t+3+16t-20=0$$ $$29t-29=0$$ $$t=1.$$ Hence $$M=(1+2,-1+3,4)=(3,2,4).$$ --- 4. **Find the reflected point $P$** Since $M$ is the midpoint of $PQ$, $$P=2M-Q.$$ Thus $$P=2(3,2,4)-(7,-2,5)=(6,4,8)-(7,-2,5)=(-1,6,3).$$ So, $$P=(-1,6,3).$$ --- 5. **Find point $R(5,p,q)$ on the line** Since $R$ lies on $L$ and its $x$-coordinate is $5$, $$1+2t=5 \implies 2t=4 \implies t=2.$$ Therefore $$p=-1+3(2)=5, \qquad q=4(2)=8.$$ So, $$R=(5,5,8).$$ --- 6. **Area of triangle $PQR$** The area of triangle $PQR$ is $$\text{Area}=\frac12\left|\overrightarrow{PQ}\times \overrightarrow{PR}\right|.$$ First find the vectors: $$\overrightarrow{PQ}=Q-P=(7-(-1),-2-6,5-3)=(8,-8,2),$$ $$\overrightarrow{PR}=R-P=(5-(-1),5-6,8-3)=(6,-1,5).$$ Now compute the cross product: $$\overrightarrow{PQ}\times \overrightarrow{PR}= \begin{vmatrix} \hat i & \hat j & \hat k \\ 8 & -8 & 2 \\ 6 & -1 & 5 \end{vmatrix}.$$ So, $$=\hat i\big((-8)(5)-2(-1)\big)-\hat j\big(8(5)-2(6)\big)+\hat k\big(8(-1)-(-8)(6)\big)$$ $$=\hat i(-40+2)-\hat j(40-12)+\hat k(-8+48)$$ $$=(-38,-28,40).$$ Hence, $$\left|\overrightarrow{PQ}\times \overrightarrow{PR}\right|^2=(-38)^2+(-28)^2+40^2$$ $$=1444+784+1600=3828.$$ Therefore, $$\text{Area}^2=\frac14\cdot 3828=957.$$ --- 7. **Final answer** $$\boxed{957}$$ The derived answer matches the stored correct answer.

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View written solutionFree

Correct answer: 957

Interpretation of “image of the point in the line”

The image of point $Q$ in the line $L$ means the reflection of $Q$ about the line $L$ .

So if $M$ is the foot of the perpendicular from $Q$ to the line $L$ , then $M$ is the midpoint of $PQ$ and $PM = QM, \quad P = 2M - Q.$

Parametric form of the line

Given $\frac{x-1}{2}=\frac{y+1}{3}=\frac{z}{4}=t,$ so a general point on $L$ is

-1+3t, 4t).$$ The line passes through $$A_0=(1,-1,0)$$ and has direction vector $$\vec d=(2,3,4).$$ Point $$Q=(7,-2,5).$$ --- 3. **Find the foot of perpendicular $M$ from $Q$ to $L$** Let $$M=(1+2t,-1+3t,4t).$$ Then $$\overrightarrow{QM}=M-Q=(1+2t-7,-1+3t+2,4t-5)=(2t-6,3t+1,4t-5).$$ Since $QM \perp L$, we must have $$\overrightarrow{QM}\cdot \vec d=0.$$ So, $$(2t-6)\cdot 2+(3t+1)\cdot 3+(4t-5)\cdot 4=0.$$ Compute: $$4t-12+9t+3+16t-20=0$$ $$29t-29=0$$ $$t=1.$$ Hence $$M=(1+2,-1+3,4)=(3,2,4).$$ --- 4. **Find the reflected point $P$** Since $M$ is the midpoint of $PQ$, $$P=2M-Q.$$ Thus $$P=2(3,2,4)-(7,-2,5)=(6,4,8)-(7,-2,5)=(-1,6,3).$$ So, $$P=(-1,6,3).$$ --- 5. **Find point $R(5,p,q)$ on the line** Since $R$ lies on $L$ and its $x$-coordinate is $5$, $$1+2t=5 \implies 2t=4 \implies t=2.$$ Therefore $$p=-1+3(2)=5, \qquad q=4(2)=8.$$ So, $$R=(5,5,8).$$ --- 6. **Area of triangle $PQR$** The area of triangle $PQR$ is $$\text{Area}=\frac12\left|\overrightarrow{PQ}\times \overrightarrow{PR}\right|.$$ First find the vectors: $$\overrightarrow{PQ}=Q-P=(7-(-1),-2-6,5-3)=(8,-8,2),$$ $$\overrightarrow{PR}=R-P=(5-(-1),5-6,8-3)=(6,-1,5).$$ Now compute the cross product: $$\overrightarrow{PQ}\times \overrightarrow{PR}= \begin{vmatrix} \hat i & \hat j & \hat k \\ 8 & -8 & 2 \\ 6 & -1 & 5 \end{vmatrix}.$$ So, $$=\hat i\big((-8)(5)-2(-1)\big)-\hat j\big(8(5)-2(6)\big)+\hat k\big(8(-1)-(-8)(6)\big)$$ $$=\hat i(-40+2)-\hat j(40-12)+\hat k(-8+48)$$ $$=(-38,-28,40).$$ Hence, $$\left|\overrightarrow{PQ}\times \overrightarrow{PR}\right|^2=(-38)^2+(-28)^2+40^2$$ $$=1444+784+1600=3828.$$ Therefore, $$\text{Area}^2=\frac14\cdot 3828=957.$$ --- 7. **Final answer** $$\boxed{957}$$ The derived answer matches the stored correct answer.

3D Geometry question

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