Let P(\alpha, \beta, \gamma) be the image of the point Q(1,6,4) in the line \fracx1=\fracy-12=\fracz-23 . Then 2 \alpha+\beta+\gamma is equal to \underline\hspace2cm

3D Geometry PYQ — JEE Main 2024 · 8 Apr · Shift 2 · Q55 with Solution

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Correct answer: 11

Interpret the line

The given line is

\frac{x}{1}=\frac{y-1}{2}=\frac{z-2}{3}.

Let the common value be $t$ . Then a parametric form of the line is

x=t,\qquad y=1+2t,\qquad z=2+3t.

So a point on the line is

A(0,1,2)

and a direction vector is

\vec d=(1,2,3).

Use the fact that image in a line means reflection about the line

If $P(\alpha,\beta,\gamma)$ is the image of $Q(1,6,4)$ in the line, then the given line is the perpendicular bisector line of segment $QP$ in 3D.

So if $R$ is the foot of the perpendicular from $Q$ to the line, then $R$ is also the midpoint of $Q$ and $P$ .

Hence,

P=2R-Q.

Find the foot of perpendicular from $Q$ to the line

A general point on the line is

R(t)=(t,1+2t,2+3t).

Since $R$ is the foot of the perpendicular from $Q(1,6,4)$ to the line, the vector

\overrightarrow{QR}=R-Q=(t-1,2t-5,3t-2)

must be perpendicular to the direction vector $(1,2,3)$ .

Therefore,

(t-1)\cdot 1+(2t-5)\cdot 2+(3t-2)\cdot 3=0.

Now simplify:

t-1+4t-10+9t-6=0

14t-17=0

t=\frac{17}{14}.

So the foot point is

R=\left(\frac{17}{14},\,1+2\cdot \frac{17}{14},\,2+3\cdot \frac{17}{14}\right).

Compute coordinates:

R=\left(\frac{17}{14},\,\frac{24}{7},\,\frac{79}{14}\right).

Find the reflected point $P$

Using

P=2R-Q,

we get

P=\left(2\cdot \frac{17}{14}-1,\,2\cdot \frac{24}{7}-6,\,2\cdot \frac{79}{14}-4\right).

So,

\alpha=\frac{17}{7}-1=\frac{10}{7},

\beta=\frac{48}{7}-6=\frac{6}{7},

\gamma=\frac{79}{7}-4=\frac{51}{7}.

Thus,

P\left(\frac{10}{7},\frac{6}{7},\frac{51}{7}\right).

Compute $2\alpha+\beta+\gamma$

2\alpha+\beta+\gamma =2\cdot \frac{10}{7}+\frac{6}{7}+\frac{51}{7} =\frac{20+6+51}{7} =\frac{77}{7}=11.

Comparison with stored answer

Our derived answer is

11.

This matches the stored correct answer.

More from 3D Geometry

View written solutionFree

Correct answer: 11

Interpret the line

The given line is

\frac{x}{1}=\frac{y-1}{2}=\frac{z-2}{3}.

Let the common value be $t$ . Then a parametric form of the line is

x=t,\qquad y=1+2t,\qquad z=2+3t.

So a point on the line is

A(0,1,2)

and a direction vector is

\vec d=(1,2,3).

Use the fact that image in a line means reflection about the line

If $P(\alpha,\beta,\gamma)$ is the image of $Q(1,6,4)$ in the line, then the given line is the perpendicular bisector line of segment $QP$ in 3D.

So if $R$ is the foot of the perpendicular from $Q$ to the line, then $R$ is also the midpoint of $Q$ and $P$ .

Hence,

P=2R-Q.

Find the foot of perpendicular from $Q$ to the line

A general point on the line is

R(t)=(t,1+2t,2+3t).

Since $R$ is the foot of the perpendicular from $Q(1,6,4)$ to the line, the vector

\overrightarrow{QR}=R-Q=(t-1,2t-5,3t-2)

must be perpendicular to the direction vector $(1,2,3)$ .

Therefore,

(t-1)\cdot 1+(2t-5)\cdot 2+(3t-2)\cdot 3=0.

Now simplify:

t-1+4t-10+9t-6=0

14t-17=0

t=\frac{17}{14}.

So the foot point is

R=\left(\frac{17}{14},\,1+2\cdot \frac{17}{14},\,2+3\cdot \frac{17}{14}\right).

Compute coordinates:

R=\left(\frac{17}{14},\,\frac{24}{7},\,\frac{79}{14}\right).

Find the reflected point $P$

Using

P=2R-Q,

we get

P=\left(2\cdot \frac{17}{14}-1,\,2\cdot \frac{24}{7}-6,\,2\cdot \frac{79}{14}-4\right).

So,

\alpha=\frac{17}{7}-1=\frac{10}{7},

\beta=\frac{48}{7}-6=\frac{6}{7},

\gamma=\frac{79}{7}-4=\frac{51}{7}.

Thus,

P\left(\frac{10}{7},\frac{6}{7},\frac{51}{7}\right).

Compute $2\alpha+\beta+\gamma$

2\alpha+\beta+\gamma =2\cdot \frac{10}{7}+\frac{6}{7}+\frac{51}{7} =\frac{20+6+51}{7} =\frac{77}{7}=11.

Comparison with stored answer

Our derived answer is

11.

This matches the stored correct answer.

3D Geometry question

3D Geometry question

More from 3D Geometry

3D Geometry question

3D Geometry question

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