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  • If the foot of the perpendicular drawn from the point (1,0,3) on line passing through  is <img alt="\left ( \frac{5}{3},\frac{7}{3},\fra

If the foot of the perpendicular drawn from the point (1,0,3) on line passing through (\alpha ,7,1) is \left ( \frac{5}{3},\frac{7}{3},\frac{17}{3} \right ), then \alphais equal to _______.
Option: 1 2
Option: 2 4
Option: 3 8
Option: 4 10
 

Answers (1)

best_answer

 

 

Angle Between Two Lines -

Vector Form
Condition for Perpendicularity

The lines are perpendicular then cos? = 90o

\\\teaxt{i.e.}\mathrm{\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;}\vec{\mathbf b}\cdot\vec{\mathbf b}'=0\;\;\;\;\;\;\;\;\;\;\;\;\; \left [\because\;\; \cos90^\circ=0 \right ]\\\\\Rightarrow\;\;\;\;a_1a_2+b_1b_2+c_1c_2=0


Condition for parallelism

The lines are parallel then \vec{\mathbf b}=\lambda\vec{\mathbf b}' for some scalar λ.

\Rightarrow \;\;\;\;\;\;\;\;\;\;\;\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}

-

Since PQ is perpendicular to L, therefore

\\\left ( 1-\frac{5}{3} \right )\left ( \alpha-\frac{5}{3} \right )+\left ( -\frac{7}{3} \right )\left ( 7-\frac{7}{3} \right )+\left ( 3-\frac{17}{3} \right )\left ( 1-\frac{17}{3} \right )=0\\\alpha=4

Posted by

Ritika Jonwal

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