Let A be the point of intersection of the lines L_1: \fracx-71=\fracy-50=\fracz-3-1 and L_2: \fracx-13=\fracy+34=\fracz+75 . Let B and C be the points on the lines L_1 and L_2 respectively such that A B=A C=\sqrt15 . Then the square of the area of the triangle A B C is :

3D Geometry PYQ — JEE Main 2025 · 4 Apr · Shift 2 · Q39 with Solution

View written solutionFree

Correct answer: D

Write the lines in parametric form

For $L_1$ : $\frac{x-7}{1}=\frac{y-5}{0}=\frac{z-3}{-1}=t$ So, $x=7+t,\quad y=5,\quad z=3-t$ Hence a direction vector of $L_1$ is $\vec d_1=(1,0,-1).$

For $L_2$ : $\frac{x-1}{3}=\frac{y+3}{4}=\frac{z+7}{5}=s$ So, $x=1+3s,\quad y=-3+4s,\quad z=-7+5s$ Hence a direction vector of $L_2$ is $\vec d_2=(3,4,5).$

Find the intersection point $A$

At the intersection, coordinates from both lines are equal.

From $y$ : $5=-3+4s \implies 4s=8 \implies s=2.$

Then from $L_2$ : $x=1+3(2)=7, \quad z=-7+5(2)=3.$

Now on $L_1$ : $x=7+t=7 \implies t=0,$ which also gives $z=3-t=3.$

Therefore, $A=(7,5,3).$

Find points $B$ and $C$ such that $AB=AC=\sqrt{15}$

Since $B$ lies on $L_1$ and $A$ is on $L_1$ , vector $\overrightarrow{AB}$ is along $\vec d_1$ .

Magnitude of $\vec d_1$ : $|\vec d_1|=\sqrt{1^2+0^2+(-1)^2}=\sqrt{2}.$

If parameter changes by $\lambda$ , then distance from $A$ is $|\lambda|\sqrt{2}.$ We need

\implies \lambda^2=\frac{15}{2}.$$ So, $$\overrightarrow{AB}=\lambda(1,0,-1), \qquad |\overrightarrow{AB}|=\sqrt{15}.$$ Similarly for $C$ on $L_2$, let $$\overrightarrow{AC}=\mu(3,4,5).$$ Magnitude of $\vec d_2$: $$|\vec d_2|=\sqrt{3^2+4^2+5^2}=\sqrt{50}=5\sqrt{2}.$$ Given $AC=\sqrt{15}$, $$|\mu|\cdot 5\sqrt{2}=\sqrt{15} \implies \mu^2=\frac{15}{50}=\frac{3}{10}.$$ --- 4. **Use area formula for triangle $ABC$** Area of triangle: $$\Delta=\frac12 |\overrightarrow{AB}\times \overrightarrow{AC}|.$$ Thus, $$\Delta^2=\frac14 |\overrightarrow{AB}\times \overrightarrow{AC}|^2.$$ Now, $$\overrightarrow{AB}=\lambda \vec d_1, \qquad \overrightarrow{AC}=\mu \vec d_2,$$ so $$|\overrightarrow{AB}\times \overrightarrow{AC}|^2=\lambda^2\mu^2 |\vec d_1\times \vec d_2|^2.$$ Compute cross product: $$\vec d_1\times \vec d_2= \begin{vmatrix} \hat i & \hat j & \hat k\\ 1 & 0 & -1\\ 3 & 4 & 5 \end{vmatrix}$$ $$=\hat i(0\cdot 5-(-1)\cdot 4)-\hat j(1\cdot 5-(-1)\cdot 3)+\hat k(1\cdot 4-0\cdot 3)$$ $$=(4,-8,4).$$ Therefore, $$|\vec d_1\times \vec d_2|^2=4^2+(-8)^2+4^2=16+64+16=96.$$ Also, $$\lambda^2\mu^2=\frac{15}{2}\cdot \frac{3}{10}=\frac{9}{4}.$$ Hence, $$|\overrightarrow{AB}\times \overrightarrow{AC}|^2=\frac{9}{4}\cdot 96=216.$$ So, $$\Delta^2=\frac14\cdot 216=54.$$ --- 5. **Final answer** The square of the area of triangle $ABC$ is $$\boxed{54}.$$ So the correct option is **D**.

More from 3D Geometry

View written solutionFree

Correct answer: D

Write the lines in parametric form

For $L_1$ : $\frac{x-7}{1}=\frac{y-5}{0}=\frac{z-3}{-1}=t$ So, $x=7+t,\quad y=5,\quad z=3-t$ Hence a direction vector of $L_1$ is $\vec d_1=(1,0,-1).$

For $L_2$ : $\frac{x-1}{3}=\frac{y+3}{4}=\frac{z+7}{5}=s$ So, $x=1+3s,\quad y=-3+4s,\quad z=-7+5s$ Hence a direction vector of $L_2$ is $\vec d_2=(3,4,5).$

Find the intersection point $A$

At the intersection, coordinates from both lines are equal.

From $y$ : $5=-3+4s \implies 4s=8 \implies s=2.$

Then from $L_2$ : $x=1+3(2)=7, \quad z=-7+5(2)=3.$

Now on $L_1$ : $x=7+t=7 \implies t=0,$ which also gives $z=3-t=3.$

Therefore, $A=(7,5,3).$

Find points $B$ and $C$ such that $AB=AC=\sqrt{15}$

Since $B$ lies on $L_1$ and $A$ is on $L_1$ , vector $\overrightarrow{AB}$ is along $\vec d_1$ .

Magnitude of $\vec d_1$ : $|\vec d_1|=\sqrt{1^2+0^2+(-1)^2}=\sqrt{2}.$

If parameter changes by $\lambda$ , then distance from $A$ is $|\lambda|\sqrt{2}.$ We need

\implies \lambda^2=\frac{15}{2}.$$ So, $$\overrightarrow{AB}=\lambda(1,0,-1), \qquad |\overrightarrow{AB}|=\sqrt{15}.$$ Similarly for $C$ on $L_2$, let $$\overrightarrow{AC}=\mu(3,4,5).$$ Magnitude of $\vec d_2$: $$|\vec d_2|=\sqrt{3^2+4^2+5^2}=\sqrt{50}=5\sqrt{2}.$$ Given $AC=\sqrt{15}$, $$|\mu|\cdot 5\sqrt{2}=\sqrt{15} \implies \mu^2=\frac{15}{50}=\frac{3}{10}.$$ --- 4. **Use area formula for triangle $ABC$** Area of triangle: $$\Delta=\frac12 |\overrightarrow{AB}\times \overrightarrow{AC}|.$$ Thus, $$\Delta^2=\frac14 |\overrightarrow{AB}\times \overrightarrow{AC}|^2.$$ Now, $$\overrightarrow{AB}=\lambda \vec d_1, \qquad \overrightarrow{AC}=\mu \vec d_2,$$ so $$|\overrightarrow{AB}\times \overrightarrow{AC}|^2=\lambda^2\mu^2 |\vec d_1\times \vec d_2|^2.$$ Compute cross product: $$\vec d_1\times \vec d_2= \begin{vmatrix} \hat i & \hat j & \hat k\\ 1 & 0 & -1\\ 3 & 4 & 5 \end{vmatrix}$$ $$=\hat i(0\cdot 5-(-1)\cdot 4)-\hat j(1\cdot 5-(-1)\cdot 3)+\hat k(1\cdot 4-0\cdot 3)$$ $$=(4,-8,4).$$ Therefore, $$|\vec d_1\times \vec d_2|^2=4^2+(-8)^2+4^2=16+64+16=96.$$ Also, $$\lambda^2\mu^2=\frac{15}{2}\cdot \frac{3}{10}=\frac{9}{4}.$$ Hence, $$|\overrightarrow{AB}\times \overrightarrow{AC}|^2=\frac{9}{4}\cdot 96=216.$$ So, $$\Delta^2=\frac14\cdot 216=54.$$ --- 5. **Final answer** The square of the area of triangle $ABC$ is $$\boxed{54}.$$ So the correct option is **D**.

3D Geometry question

3D Geometry question

More from 3D Geometry

3D Geometry question

3D Geometry question

More from 3D Geometry