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Equation of the straight line in the plane \overline{r}.\overline{n}=d which is parallel to \overline{r}=\overline{a}+\lambda \overline{b} and passes through the foot of perpendicular drawn from the point p\left ( \overline{a} \right )to the \overline{r}.\overline{n}=d is (where \overline{r}.\overline{b}=0)

Option: 1

\vec{r}=\vec{a}+\frac{(d-\vec{a}.\vec{n})}{\left | \vec{n} \right |^{2}}\vec{n}+\lambda \vec{b}


Option: 2

\overline{r}=\overline{a}+\left ( \frac{d-\overline{a}.\overline{n}}{\overline{n}} \right )\overline{n}+\lambda \overline{b}


Option: 3

\overline{r}=\overline{a}+\left ( \frac{\overline{a}.\overline{n}-d}{\overline{n}^{2}} \right )\overline{n}+\lambda \overline{b}


Option: 4

\overline{r}=\overline{a}+\left ( \frac{\overline{a}.\overline{n}-d}{\overline{n}} \right )\overline{n}+\lambda \overline{b}


Answers (1)

best_answer

 

Foot of perpendicular -

Let L be the foot of perpendicular drawn from P\left ( \alpha ,\beta ,\gamma \right ) on the plane ax+by+cz+d=0

L will be given by the formula

\frac{x-\alpha }{a}=\frac{y-\beta }{b}=\frac{z-\gamma }{c}= \frac{-\left ( a\alpha +b\beta +c\gamma +d \right )}{a^{2}+b^{2}+c^{2}}

-

 

 

Foot of perpendicular from point  A(\vec{a})  on the plane \vec{r}.\vec{n}=d is \vec{a}+\frac{(d-\vec{a}.\vec{n})}{\left | \vec{n} \right |^{2}}\vec{n}

\therefore Equation of line parallel to \vec{r}=\vec{a}+\lambda \vec{b} in the plane \vec{r}.\vec{n}=d is given by \vec{r}=\vec{a}+\frac{(d-\vec{a}.\vec{n})}{\left | \vec{n} \right |^{2}}\vec{n}+\lambda \vec{b}

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Sayak

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