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If the angle between the line x=\frac{y-1}{2}=\frac{z-3}{\lambda }   and  the plane x+2y+3z=4\; is\; \cos ^{-1}\left ( \sqrt{\frac{5}{14}} \right ),then\; \lambda \; equals

  • Option 1)

    2/5

  • Option 2)

    5/3

  • Option 3)

    2/3

  • Option 4)

    3/2

 

Answers (1)

best_answer

As learnt in

Angle between two lines (Vector form ) -

Let the two lines be \vec{r}= \vec{a}+\lambda \vec{b}\, and\, \vec{r}= \vec{a_{1}}+\lambda \vec{b_{1}} .The angle between two lines will be equal to angle between their parallel vectors \vec{b}\, and \, \vec{b_{1}} .

\cos \Theta =\frac{\vec{b}\cdot \vec{b_{1}}}{\left | \vec{b} \right |\left | \vec{b_{1}} \right |}

-

 

 

Projection of a line segment on a line -

Projection of line segment joining the points P(x1,y1,z1) and Q(x2,y2,z2) on a line having direction cosines (l,m,n) is

\left | \left ( x_{2}-x_{1} \right )l+\left ( y_{2}-y_{1} \right )m+\left ( z_{2}-z_{1} \right )n\right |

- wherein

 \cos \theta= \frac{\sqrt{5}}{\sqrt{14}}

Then, \sin \theta= \frac{3}{\sqrt{14}}

So, \frac{3}{\sqrt{14}}= \frac{1+4+3\lambda }{\sqrt{5+\lambda ^{2}}\sqrt{14}}

9({5+\lambda ^{2}})= (3\lambda +5)^{2}

\Rightarrow \lambda = \frac{2}{3}

 

 

 


Option 1)

2/5

Incorrect option

Option 2)

5/3

Incorrect option

Option 3)

2/3

Correct option

Option 4)

3/2

Incorrect option

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divya.saini

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