Let the image of the point (1,0,7) in the line \fracx1=\fracy-12=\fracz-23 be the point (\alpha, \beta, \gamma) . Then which one of the following points lies on the line passing through (\alpha, \beta, \gamma) and making angles \frac2 \pi3 and \frac3 \pi4 with y -axis and z -axis respectively and an acute angle with x -axis ?

3D Geometry PYQ — JEE Main 2024 · 27 Jan · Shift 2 · Q35 with Solution

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Correct answer: C

Interpret the given line

The line is

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

So its parametric form is

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

Hence, a point on the line is $A=(0,1,2),$ and a direction vector is $\vec d=(1,2,3).$

Find the image of point $P=(1,0,7)$ in the line

“Image in the line” means reflection of the point about the line.

First, find the foot of perpendicular from $P$ to the line. Let the foot be $Q=(t,1+2t,2+3t).$ Then $\overrightarrow{AQ}=t(1,2,3),$ and $\overrightarrow{PQ}=Q-P=(t-1,1+2t,3t-5).$ For perpendicularity,

So,

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

t-1+2+4t+9t-15=0

14t-14=0

t=1.

Thus, $Q=(1,3,5).$

Now $Q$ is the midpoint of $P=(1,0,7)$ and its image $P'=(\alpha,\beta,\gamma)$ . Hence,

P'=2Q-P=(2,6,10)-(1,0,7)=(1,6,3).

Therefore,

(\alpha,\beta,\gamma)=(1,6,3).

Find direction ratios of the required line

The line through $(\alpha,\beta,\gamma)=(1,6,3)$ makes angles:

$\frac{2\pi}{3}$ with the $y$ -axis,
$\frac{3\pi}{4}$ with the $z$ -axis,
an acute angle with the $x$ -axis.

Let its direction cosines be $(l,m,n)$ . Then

m=\cos\frac{2\pi}{3}=-\frac12,

n=\cos\frac{3\pi}{4}=-\frac{1}{\sqrt2}.

Using $l^2+m^2+n^2=1,$ we get

l^2+\left(-\frac12\right)^2+\left(-\frac1{\sqrt2}\right)^2=1

l^2+\frac14+\frac12=1

l^2=\frac14.

So $l=\pm \frac12.$ Since the angle with the $x$ -axis is acute, $l>0$ . Hence,

Therefore direction cosines are $\left(\frac12,-\frac12,-\frac1{\sqrt2}\right).$ A proportional direction vector is $(1,-1,-\sqrt2).$

So the required line is

(x,y,z)=(1,6,3)+\lambda(1,-1,-\sqrt2).

Check which option lies on this line

For a point on the line,

Option A: $(1,-2,1+\sqrt2)$

From $x=1$ , we get $\lambda=0$ . Then $y=6\ne -2$ . So A does not lie on the line.

Option B: $(3,-4,3+2\sqrt2)$

From $x=3$ , we get $\lambda=2$ . Then $y=6-2=4\ne -4$ . So B does not lie on the line.

Option C: $(3,4,3-2\sqrt2)$

From $x=3$ , we get $\lambda=2$ . Then $y=6-2=4,$ $z=3-2\sqrt2,$ which matches exactly. So C lies on the line.

Option D: $(1,2,1-\sqrt2)$

From $x=1$ , we get $\lambda=0$ . Then $y=6\ne 2$ . So D does not lie on the line.

Conclusion

The correct option is $\boxed{\text{C } (3,4,3-2\sqrt2)}.$

More from 3D Geometry

View written solutionFree

Correct answer: C

Interpret the given line

The line is

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

So its parametric form is

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

Hence, a point on the line is $A=(0,1,2),$ and a direction vector is $\vec d=(1,2,3).$

Find the image of point $P=(1,0,7)$ in the line

“Image in the line” means reflection of the point about the line.

So,

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

t-1+2+4t+9t-15=0

14t-14=0

t=1.

Thus, $Q=(1,3,5).$

Now $Q$ is the midpoint of $P=(1,0,7)$ and its image $P'=(\alpha,\beta,\gamma)$ . Hence,

P'=2Q-P=(2,6,10)-(1,0,7)=(1,6,3).

Therefore,

(\alpha,\beta,\gamma)=(1,6,3).

Find direction ratios of the required line

The line through $(\alpha,\beta,\gamma)=(1,6,3)$ makes angles:

$\frac{2\pi}{3}$ with the $y$ -axis,
$\frac{3\pi}{4}$ with the $z$ -axis,
an acute angle with the $x$ -axis.

Let its direction cosines be $(l,m,n)$ . Then

m=\cos\frac{2\pi}{3}=-\frac12,

n=\cos\frac{3\pi}{4}=-\frac{1}{\sqrt2}.

Using $l^2+m^2+n^2=1,$ we get

l^2+\left(-\frac12\right)^2+\left(-\frac1{\sqrt2}\right)^2=1

l^2+\frac14+\frac12=1

l^2=\frac14.

So $l=\pm \frac12.$ Since the angle with the $x$ -axis is acute, $l>0$ . Hence,

Therefore direction cosines are $\left(\frac12,-\frac12,-\frac1{\sqrt2}\right).$ A proportional direction vector is $(1,-1,-\sqrt2).$

So the required line is

(x,y,z)=(1,6,3)+\lambda(1,-1,-\sqrt2).

Check which option lies on this line

For a point on the line,

Option A: $(1,-2,1+\sqrt2)$

From $x=1$ , we get $\lambda=0$ . Then $y=6\ne -2$ . So A does not lie on the line.

Option B: $(3,-4,3+2\sqrt2)$

From $x=3$ , we get $\lambda=2$ . Then $y=6-2=4\ne -4$ . So B does not lie on the line.

Option C: $(3,4,3-2\sqrt2)$

From $x=3$ , we get $\lambda=2$ . Then $y=6-2=4,$ $z=3-2\sqrt2,$ which matches exactly. So C lies on the line.

Option D: $(1,2,1-\sqrt2)$

From $x=1$ , we get $\lambda=0$ . Then $y=6\ne 2$ . So D does not lie on the line.

Conclusion

The correct option is $\boxed{\text{C } (3,4,3-2\sqrt2)}.$

3D Geometry question

3D Geometry question

Option A: $(1,-2,1+\sqrt2)$

Option B: $(3,-4,3+2\sqrt2)$

Option C: $(3,4,3-2\sqrt2)$

Option D: $(1,2,1-\sqrt2)$

More from 3D Geometry

3D Geometry question

3D Geometry question

Option A: $(1,-2,1+\sqrt2)$

Option B: $(3,-4,3+2\sqrt2)$

Option C: $(3,4,3-2\sqrt2)$

Option D: $(1,2,1-\sqrt2)$

More from 3D Geometry

Option A: (1,−2,1+2)(1,-2,1+\sqrt2)(1,−2,1+2​)

Option B: (3,−4,3+22)(3,-4,3+2\sqrt2)(3,−4,3+22​)

Option C: (3,4,3−22)(3,4,3-2\sqrt2)(3,4,3−22​)

Option D: (1,2,1−2)(1,2,1-\sqrt2)(1,2,1−2​)

More from 3D Geometry

Option A: (1,−2,1+2)(1,-2,1+\sqrt2)(1,−2,1+2​)

Option B: (3,−4,3+22)(3,-4,3+2\sqrt2)(3,−4,3+22​)

Option C: (3,4,3−22)(3,4,3-2\sqrt2)(3,4,3−22​)

Option D: (1,2,1−2)(1,2,1-\sqrt2)(1,2,1−2​)

More from 3D Geometry

Option A: $(1,-2,1+\sqrt2)$

Option B: $(3,-4,3+2\sqrt2)$

Option C: $(3,4,3-2\sqrt2)$

Option D: $(1,2,1-\sqrt2)$

Option A: $(1,-2,1+\sqrt2)$

Option B: $(3,-4,3+2\sqrt2)$

Option C: $(3,4,3-2\sqrt2)$

Option D: $(1,2,1-\sqrt2)$