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  • Answer please! - Three Dimensional Geometry - JEE Main-4

Foot of perpendicular drawn from \hat{i}+\hat{j}+\hat{k} to the line \vec{r}=(\hat{i}+2\hat{j}+3\hat{k}) + \lambda(2\hat{i}+3\hat{j}+4\hat{k}) will have position vector

  • Option 1)

    \frac{1}{29}(7\hat{i}+25\hat{j}+43\hat{k})

  • Option 2)

    \frac{1}{29}(7\hat{i}-25\hat{j}+43\hat{k})

  • Option 3)

    \frac{1}{29}(7\hat{i}+25\hat{j}-43\hat{k})

  • Option 4)

    \frac{1}{29}(7\hat{i}-25\hat{j}-43\hat{k})

 

Answers (1)

best_answer

As we have learned

Perpendicular distance of a point from a line, foot of perpendicular and image (vector form ) -

Foot of perpendicular is \vec{a}-\left ( \frac{\left ( \vec{a}-\vec{\alpha } \right )\cdot \vec{b}}{\left | b \right |^{2}} \right )\vec{b}

Perpendicular distance =\left | \vec{a}-\left ( \frac{\left ( \vec{a}-\vec{\alpha } \right )\cdot \vec{b}}{\left | b \right |^{2}} \right )\vec{b}-\vec{\alpha } \right |

image is 2\vec{a}-\left ( \frac{2\left ( \vec{a}-\vec{\alpha } \right )\cdot \vec{b}}{\left | b \right |^{2}} \right )\vec{b}-\vec{\alpha }    

- wherein

Let L be the foot of perpendicular drawn from P\left ( \vec{\alpha } \right ) on the line \vec{r}= \vec{a}+\lambda \vec{b}

Since\vec{r} denotes the position vector of any point on the line therefore position vector of L will be  \vec{a}+\lambda \vec{b}

DR’s of PL will be \vec{a}+\lambda \vec{b}-\vec{\alpha }

Now PL is perpendicular to \vec{b}

\left ( \vec{a}+\lambda \vec{b}-\vec{\alpha } \right )\cdot \vec{b}= 0

\lambda = \frac{-\left ( \vec{a}-\vec{\alpha } \right )\cdot \vec{b}}{\left | b \right |^{2}},position vector of L will be \vec{a}-\left ( \frac{\left ( \vec{a}-\vec{\alpha } \right )\cdot \vec{b}}{\left | b \right |^{2}} \right )\vec{b}

the distance between L and P is perpendicular distance

Let the image be Q\left ( \vec{\beta } \right ) then \frac{\vec{\alpha }+\vec{\beta }}{2}=\vec{a}-\left ( \frac{\left ( \vec{a}-\vec{\alpha } \right )\cdot \vec{b}}{\left | b \right |^{2}} \right )\vec{b}

\vec{\beta }= 2\vec{a}-\left ( \frac{2\left ( \vec{a}-\vec{\alpha } \right )\cdot \vec{b}}{\left | b \right |^{2}} \right )\vec{b}-\vec{\alpha }

 

 

 

 here\vec{a}=\hat{i}+2\hat{j}+3\hat{k}, \vec{b}=2\hat{i}+3\hat{j}+9\hat{k}

\vec{\alpha}=\hat{i}+\hat{j}+\hat{k}

\therefore Foot of perpendicular =\vec{a}-(\frac{(\vec{a}-\vec{\alpha})\cdot\vec{b})}{\left|\vec{b}^{2}\right|})\vec{b}

\vec{a}-\vec{\alpha}=\hat{j}+2\hat{k} \Rightarrow (\vec{a}-\vec{\alpha})\cdot\vec{b}=11

\therefore Foot of perpendicular =\(\hat{i}+2\hat{j}+3\hat{k})-\frac{11(2\hat{i}+3\hat{j}+4\hat{k})}{29}

=\frac{1}{29}(7\hat{i}+25\hat{j}+43\hat{k})

\therefore Option (A)


Option 1)

\frac{1}{29}(7\hat{i}+25\hat{j}+43\hat{k})

Option 2)

\frac{1}{29}(7\hat{i}-25\hat{j}+43\hat{k})

Option 3)

\frac{1}{29}(7\hat{i}+25\hat{j}-43\hat{k})

Option 4)

\frac{1}{29}(7\hat{i}-25\hat{j}-43\hat{k})

Posted by

Himanshu

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