Consider a line L passing through the points P(1,2,1) and Q(2,1,-1) . If the mirror image of the point A(2,2,2) in the line L is (\alpha, \beta, \gamma) , then \alpha+\beta+6 \gamma is equal to \underline\hspace2cm .

3D Geometry PYQ — JEE Main 2024 · 4 Apr · Shift 2 · Q59 with Solution

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Correct answer: 6

Equation of the line $L$

The line passes through $P(1,2,1), \, Q(2,1,-1).$ So its direction vector is $\vec d = Q-P = (2-1,1-2,-1-1)=(1,-1,-2).$

Hence the line is $\mathbf r = (1,2,1) + t(1,-1,-2).$

Find the foot of perpendicular from $A(2,2,2)$ to the line

Let the foot of perpendicular be $H$ on the line, so $H=(1+t,\,2-t,\,1-2t).$

Since $AH \perp L$ , we must have $\overrightarrow{AH}\cdot \vec d = 0.$

Now, $\overrightarrow{AH}=H-A=((1+t)-2,(2-t)-2,(1-2t)-2)=(t-1,-t,-1-2t).$

Dot product with $(1,-1,-2)$ :

(t-1)(1)+(-t)(-1)+(-1-2t)(-2)=0.

Simplifying, $t-1+t+2+4t=0$ $6t+1=0$ $t=-\frac16.$

Therefore, $H=\left(1-\frac16,\,2+\frac16,\,1+\frac13\right)=\left(\frac56,\,\frac{13}{6},\,\frac43\right).$

Use midpoint property of reflection

If $A'(\alpha,\beta,\gamma)$ is the mirror image of $A$ in the line $L$ , then the line $L$ is the perpendicular bisector in 3D of segment $AA'$ . Thus $H$ is the midpoint of $A$ and $A'$ .

So, $A' = 2H - A.$

Compute:

= \left(\frac53,\frac{13}{3},\frac83\right) - (2,2,2).$$ Thus, $$A' = \left(\frac53-2,\frac{13}{3}-2,\frac83-2\right) = \left(-\frac13,\frac73,\frac23\right).$$ Hence, $$\alpha=-\frac13,\quad \beta=\frac73,\quad \gamma=\frac23.$$ --- 4. **Find $\alpha+\beta+6\gamma$** $$\alpha+\beta+6\gamma = -\frac13+\frac73+6\cdot\frac23.$$ $$=\frac{6}{3}+4=2+4=6.$$ --- 5. **Comparison with stored answer** Derived answer = $6$. Stored correct answer = $6$. They agree.

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View written solutionFree

Correct answer: 6

Equation of the line $L$

The line passes through $P(1,2,1), \, Q(2,1,-1).$ So its direction vector is $\vec d = Q-P = (2-1,1-2,-1-1)=(1,-1,-2).$

Hence the line is $\mathbf r = (1,2,1) + t(1,-1,-2).$

Find the foot of perpendicular from $A(2,2,2)$ to the line

Let the foot of perpendicular be $H$ on the line, so $H=(1+t,\,2-t,\,1-2t).$

Since $AH \perp L$ , we must have $\overrightarrow{AH}\cdot \vec d = 0.$

Now, $\overrightarrow{AH}=H-A=((1+t)-2,(2-t)-2,(1-2t)-2)=(t-1,-t,-1-2t).$

Dot product with $(1,-1,-2)$ :

(t-1)(1)+(-t)(-1)+(-1-2t)(-2)=0.

Simplifying, $t-1+t+2+4t=0$ $6t+1=0$ $t=-\frac16.$

Therefore, $H=\left(1-\frac16,\,2+\frac16,\,1+\frac13\right)=\left(\frac56,\,\frac{13}{6},\,\frac43\right).$

Use midpoint property of reflection

If $A'(\alpha,\beta,\gamma)$ is the mirror image of $A$ in the line $L$ , then the line $L$ is the perpendicular bisector in 3D of segment $AA'$ . Thus $H$ is the midpoint of $A$ and $A'$ .

So, $A' = 2H - A.$

Compute:

= \left(\frac53,\frac{13}{3},\frac83\right) - (2,2,2).$$ Thus, $$A' = \left(\frac53-2,\frac{13}{3}-2,\frac83-2\right) = \left(-\frac13,\frac73,\frac23\right).$$ Hence, $$\alpha=-\frac13,\quad \beta=\frac73,\quad \gamma=\frac23.$$ --- 4. **Find $\alpha+\beta+6\gamma$** $$\alpha+\beta+6\gamma = -\frac13+\frac73+6\cdot\frac23.$$ $$=\frac{6}{3}+4=2+4=6.$$ --- 5. **Comparison with stored answer** Derived answer = $6$. Stored correct answer = $6$. They agree.

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