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The coordinates of the foot of the perpendicular from the point (1, −2, 1) on the plane containing the lines

\small \frac{x+1}{6} = \frac{y-1}{7} = \frac{z-3}{8}  and

\small \frac{x-1}{3} = \frac{y-2}{5} = \frac{z-3}{7} , is

  • Option 1)

    (2, −4, 2)

  • Option 2)

    (−1, 2, −1)

  • Option 3)

     (0, 0, 0)

  • Option 4)

    (1, 1, 1)

     

 

Answers (1)

best_answer

As we learnt in 

Cartesian equation of plane passing through a given point and normal to a given vector -

\left ( x-x_{0} \right )a+\left ( y-y_{0} \right )b+\left ( z-z_{0} \right )c= 0
 

- wherein

\vec{r}= x\hat{i}+y\hat{j}+z\hat{k}

\vec{a}= x_{0}\hat{i}+y_{0}\hat{j}+z_{0}\hat{k}

\vec{n}= a\hat{i}+b\hat{j}+c\hat{k}

Putting in

\left ( \vec{r}-\vec{a} \right )\cdot \vec{n}= 0

We get \left ( x-x_{0} \right )a+\left ( y-y_{0} \right )b+\left ( z-z_{0} \right )c= 0

 

 First, we find the equation of plane normal vector of plane containing L_{1} and L_{2} is

x^{-1}=\begin{bmatrix} \hat{i}&\hat{t} &\hat{k} \\ 6 &7 &8 \\ 3 &5 & 7 \end{bmatrix}

=9\hat{i}-18\hat{j}+9\hat{k}

Unit vector \hat{x}=\frac{\hat{i}-2\hat{j}+\hat{k}}{\sqrt{6}}

Hence, equation of plane is of the form

x-2y+z=k

It passes through (-1,1,3)

-1-2+3=k\Rightarrow 0

So plance is x-2y+z=0

Foot of perpendicular is (0,0,0)


Option 1)

(2, −4, 2)

This option is incorrect

Option 2)

(−1, 2, −1)

This option is incorrect

Option 3)

 (0, 0, 0)

This option is correct

Option 4)

(1, 1, 1)

 

This option is incorrect

Posted by

prateek

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