Question

A straight line $r=a+\lambda b$ meets the plane $r.n=0$ at $P$. Then the position vector $P$ is

• A. $a+\frac{a.n}{b.n}b$
• B. $a-\frac{a.n}{b.n}b$
• C. $a+\frac{b.n}{a.n}b$
• D. None of these

Answer 1

Answer: (B)

$a-\frac{a.n}{b.n}b$

Given eq of line and plane

$r=a+\lambda b$-------(1)

$r\cdot n=0$------(2)

the intersecting point P from eq of line be

$P(a,\lambda b)$

Here the plane also contains point P so passing through P

$n\cdot (a+\lambda b)=0$

$n\cdot a+r\cdot \lambda =0$

$\lambda n\cdot b=-n\cdot a$

$\lambda =-\frac{n\cdot a}{n\cdot b}$

putting value of $\lambda$ in point P

$P(a,-\frac{n\cdot a}{n\cdot b}b)$

position vector of P

$OP=a-\frac{n\cdot a}{n\cdot b}b$