Q

# Solve! - Three Dimensional Geometry - JEE Main

The shortest distance between the z-axis and the line

• Option 1)

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• Option 2)

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• Option 3)

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• Option 4)

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As we learnt in

Shortest distance between two skew lines (Cartesian form) -

Shortest distance between

$\frac{x-x_{1}}{a_{1}}=\frac{y-y_{1}}{b_{1}}=\frac{z-z_{1}}{c_{1}}$and $\frac{x-x_{2}}{a_{2}}=\frac{y-y_{2}}{b_{2}}=\frac{z-z_{2}}{c_{2}}$is given by

$\left | \frac{\left ( \vec{b} \times \vec{b_{1}}\right )\cdot \left ( \vec{a} -\vec{a_{1}}\right )}{\left | \vec{b} \times \vec{b_{1}} \right |} \right |$ Where

$\vec{a}= x_{1}\hat{i}+y_{1}\hat{j}+z_{1}\hat{k}$

$\vec{a_{1}}= x_{2}\hat{i}+y_{2}\hat{j}+z_{2}\hat{k}$

$\vec{b}= a_{1}\hat{i}+b_{1}\hat{j}+c_{1}\hat{k}$

$\vec{b_{1}}= a_{2}\hat{i}+b_{2}\hat{j}+c_{2}\hat{k}$

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Normal vector of line is $\begin{vmatrix} \hat{i} & \hat{j}& \hat{k}\\ 1 &1 & 2\\ 2 & 3 & 4 \end{vmatrix}=-2\hat{i}+\hat{k}$

Put z=0

We get x+y=3

2x+3y=4

y=-2, x=5

(5,-2,0) is the point

Equation of line will be $\frac{x-5}{-2}=\frac{y+2}{0}=\frac{z}{1}$

Equation of z-axis is $\frac{x}{0}=\frac{y}{0}=\frac{z}{1}$

Shortest distance = $\frac{\left | ((-2\hat{i}+\hat{k})\times \hat{k}).(5\hat{i}-2\hat{j}) \right |}{\left | (-2\hat{i}+\hat{k})\times \hat{k} \right |}$

$=\frac{4}{2}=2$

Option 1)

1

This is incorrect option

Option 2)

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This is correct option

Option 3)

3

This is incorrect option

Option 4)

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This is incorrect option

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