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The shortest distance between the z-axis and the line

x+y+2z-3=0=2x+3y+4z-4,\; is:

 

  • Option 1)

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  • Option 2)

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  • Option 3)

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  • Option 4)

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Answers (1)

best_answer

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

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Option 2)

2

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Option 3)

3

This is incorrect option

Option 4)

4

This is incorrect option

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