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# Answer please! - Three Dimensional Geometry - JEE Main-5

If the line  lies in the plane,

2x−4y+3z=2,

then the shortest distance between this line and the line

is

• Option 1)

2

• Option 2)

1

• Option 3)

0

• Option 4)

3

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As we have learned

Shortest distance between two skew lines (vector form) -

Shortest distance between $\vec{r}=\vec{a}+\lambda \vec{b}\, \: and \,\: \vec{r}=\vec{a_{1}}+\mu\vec{b}$is given by

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

- wherein

shortest distance is among the line which is perpendicular to both

$LM= \vec{b}\times \vec{b_{1}}$

shortest distance will be projection of PQ = $\vec{a}- \vec{a_{1}}$ on LM

$(3 , -2 , -\lambda ) \: \: lies \: \: on \: \: plane$

So,  $6+8 -3 \lambda = 2 \\ \Rightarrow 3 \lambda = 12 \\ \lambda = 4$

Where $\vec b \times \vec c = \begin{vmatrix} \hat i & \hat j & \hat k \\ 1 & -1 &-2 \\ 12 & 9 & 4 \end{vmatrix} = 14 \hat i -28 \hat j +21 \hat k$

So distance  = 0

Option 1)

2

Option 2)

1

Option 3)

0

Option 4)

3

As we have learned

Shortest distance between two skew lines (vector form) -

Shortest distance between $\vec{r}=\vec{a}+\lambda \vec{b}\, \: and \,\: \vec{r}=\vec{a_{1}}+\mu\vec{b}$is given by

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

- wherein

shortest distance is among the line which is perpendicular to both

$LM= \vec{b}\times \vec{b_{1}}$

shortest distance will be projection of PQ = $\vec{a}- \vec{a_{1}}$ on LM

$(3 , -2 , -\lambda ) \: \: lies \: \: on \: \: plane$

So,  $6+8 -3 \lambda = 2 \\ \Rightarrow 3 \lambda = 12 \\ \lambda = 4$

Where $\vec b \times \vec c = \begin{vmatrix} \hat i & \hat j & \hat k \\ 1 & -1 &-2 \\ 12 & 9 & 4 \end{vmatrix} = 14 \hat i -28 \hat j +21 \hat k$

So distance  = 0

Option 1)

2

Option 2)

1

Option 3)

0

Option 4)

3

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