Question

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

• A. $\overrightarrow{r}=\overrightarrow{a}+\left(\frac{d-\overrightarrow{a}\cdot \overrightarrow{n}}{n^2}\right)\overrightarrow{n}+\lambda \overrightarrow{b}$
• B. $\overrightarrow{r}=\overrightarrow{a}+\left(\frac{d-\overrightarrow{a}\cdot \overrightarrow{n}}{n}\right)\overrightarrow{n}+\lambda \overrightarrow{b}$
• C. $\overrightarrow{r}=\overrightarrow{a}+\left(\frac{\overrightarrow{a}\cdot \overrightarrow{n}-d}{n^2}\right)\overrightarrow{n}+\lambda \overrightarrow{b}$
• D. $\overrightarrow{r}=\overrightarrow{a}+\left(\frac{\overrightarrow{a}\cdot \overrightarrow{n}-d}{n}\right)\overrightarrow{n}+\lambda \overrightarrow{b}$

Answer 1

The correct option is A $\overrightarrow{r}=\overrightarrow{a}+\left(\frac{d-\overrightarrow{a}\cdot \overrightarrow{n}}{n^2}\right)\overrightarrow{n}+\lambda \overrightarrow{b}$

The equation of line passing through $\overrightarrow{a}$ and perpendicular to plane be $\overrightarrow{r}.\overrightarrow{n}=d$ is $\overrightarrow{r}=\overrightarrow{a}+t\overrightarrow{n}$
Let point of intersection of the plane with the line be $\overrightarrow{s}=\overrightarrow{a}+t\overrightarrow{n}$
Since, $S$ lies on the plane $\overrightarrow{r}.\overrightarrow{n}=d$
Therefore, $(\overrightarrow{a}+t\overrightarrow{n}).\overrightarrow{n}=d$
$\Rightarrow t=\frac{d-\overrightarrow{a}.\overrightarrow{n}}{|n{|}^2}$
$\Rightarrow \overrightarrow{s}=\overrightarrow{a}+t\overrightarrow{n}=\overrightarrow{a}+\left(\frac{(d-\overrightarrow{a}\cdot \overrightarrow{n})}{{\mid \overrightarrow{n}\mid }^2}\right)\overrightarrow{n}$
Therefore, equation of line passing through $\overrightarrow{s}$ and parallel $\overrightarrow{r}=\overrightarrow{a}+\lambda \overrightarrow{b}$ is $\overrightarrow{r}=\overrightarrow{a}+\left(\frac{(d-\overrightarrow{a}\cdot \overrightarrow{n})}{{\mid \overrightarrow{n}\mid }^2}\right)\overrightarrow{n}+\lambda \overrightarrow{b}$