If the length of the perpendicular from the point ( \beta , 0, \beta ) ( \beta e 0) to the line, x \over 1 = y - 1 \over 0 = z + 1 \over - 1 is \sqrt 3 \over 2 , then \beta is equal to :

3D Geometry PYQ — JEE Main 2019 · 10 Apr · Shift 1 · Q25 with Solution

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Correct answer: D

Interpret the line

The line is given in symmetric form: $\frac{x}{1}=\frac{y-1}{0}=\frac{z+1}{-1}$

So it passes through the point $A=(0,1,-1)$ and has direction vector $\vec d=(1,0,-1).$

Given point

The point is $P=(\beta,0,\beta).$

Use distance from a point to a line in 3D

Distance from point $P$ to the line through $A$ with direction vector $\vec d$ is $\text{Distance} = \frac{|\overrightarrow{AP}\times \vec d|}{|\vec d|}.$

First, $\overrightarrow{AP}=P-A=(\beta-0,0-1,\beta-(-1))=(\beta,-1,\beta+1).$

Now compute the cross product:

\begin{vmatrix} \hat i & \hat j & \hat k\\ \beta & -1 & \beta+1\\ 1 & 0 & -1 \end{vmatrix}$$ $$=\hat i\big((-1)(-1)- (\beta+1)\cdot 0\big) -\hat j\big(\beta(-1)- (\beta+1)(1)\big) +\hat k\big(\beta\cdot 0-(-1)(1)\big)$$ $$=\hat i(1)-\hat j(-\beta-\beta-1)+\hat k(1)$$ $$=(1,2\beta+1,1).$$ Hence, $$|\overrightarrow{AP}\times \vec d|=\sqrt{1^2+(2\beta+1)^2+1^2} =\sqrt{(2\beta+1)^2+2}.$$ Also, $$|\vec d|=\sqrt{1^2+0^2+(-1)^2}=\sqrt2.$$ Therefore distance is $$\frac{\sqrt{(2\beta+1)^2+2}}{\sqrt2}.$$ Given this equals $\sqrt{\frac32}$, so $$\frac{\sqrt{(2\beta+1)^2+2}}{\sqrt2}=\sqrt{\frac32}.$$ 4. **Solve for $\beta$** Squaring both sides: $$\frac{(2\beta+1)^2+2}{2}=\frac32$$ $$ (2\beta+1)^2+2=3$$ $$ (2\beta+1)^2=1$$ So, $$2\beta+1=\pm 1.$$ Case 1: $$2\beta+1=1 \implies 2\beta=0 \implies \beta=0.$$ Case 2: $$2\beta+1=-1 \implies 2\beta=-2 \implies \beta=-1.$$ 5. **Match with options** The possible values are $$\beta=0 \quad \text{or} \quad \beta=-1.$$ Among the given options, only $$\boxed{-1}$$ is present. So the correct option is **D**.

More from 3D Geometry

View written solutionFree

Correct answer: D

Interpret the line

The line is given in symmetric form: $\frac{x}{1}=\frac{y-1}{0}=\frac{z+1}{-1}$

So it passes through the point $A=(0,1,-1)$ and has direction vector $\vec d=(1,0,-1).$

Given point

The point is $P=(\beta,0,\beta).$

Use distance from a point to a line in 3D

Distance from point $P$ to the line through $A$ with direction vector $\vec d$ is $\text{Distance} = \frac{|\overrightarrow{AP}\times \vec d|}{|\vec d|}.$

First, $\overrightarrow{AP}=P-A=(\beta-0,0-1,\beta-(-1))=(\beta,-1,\beta+1).$

Now compute the cross product:

\begin{vmatrix} \hat i & \hat j & \hat k\\ \beta & -1 & \beta+1\\ 1 & 0 & -1 \end{vmatrix}$$ $$=\hat i\big((-1)(-1)- (\beta+1)\cdot 0\big) -\hat j\big(\beta(-1)- (\beta+1)(1)\big) +\hat k\big(\beta\cdot 0-(-1)(1)\big)$$ $$=\hat i(1)-\hat j(-\beta-\beta-1)+\hat k(1)$$ $$=(1,2\beta+1,1).$$ Hence, $$|\overrightarrow{AP}\times \vec d|=\sqrt{1^2+(2\beta+1)^2+1^2} =\sqrt{(2\beta+1)^2+2}.$$ Also, $$|\vec d|=\sqrt{1^2+0^2+(-1)^2}=\sqrt2.$$ Therefore distance is $$\frac{\sqrt{(2\beta+1)^2+2}}{\sqrt2}.$$ Given this equals $\sqrt{\frac32}$, so $$\frac{\sqrt{(2\beta+1)^2+2}}{\sqrt2}=\sqrt{\frac32}.$$ 4. **Solve for $\beta$** Squaring both sides: $$\frac{(2\beta+1)^2+2}{2}=\frac32$$ $$ (2\beta+1)^2+2=3$$ $$ (2\beta+1)^2=1$$ So, $$2\beta+1=\pm 1.$$ Case 1: $$2\beta+1=1 \implies 2\beta=0 \implies \beta=0.$$ Case 2: $$2\beta+1=-1 \implies 2\beta=-2 \implies \beta=-1.$$ 5. **Match with options** The possible values are $$\beta=0 \quad \text{or} \quad \beta=-1.$$ Among the given options, only $$\boxed{-1}$$ is present. So the correct option is **D**.

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