A vector v in the first octant is inclined to the x-axis at 60^ , to the y-axis at 45 and …

Question

A vector $\vec{v}$ in the first octant is inclined to the $x$-axis at ${60}^{\circ }$, to the $y$-axis at 45 and to the $z$-axis at an acute angle. If a plane passing through the points $(\sqrt{2},-1,1)$ and $(a,b,c)$, is normal to $\vec{v}$, then :

Accepted Answer

$$l = {1 \over 2},m = {1 \over {\sqrt 2 }},n = \cos \theta $$

${l^2} + {m^2} + {n^2} = 1$

$$ \Rightarrow {1 \over 4} + {1 \over 2} + {n^2} = 1 \Rightarrow {n^2} = {1 \over 4} \Rightarrow n = \, + \,{1 \over 2}$$

$$\theta$$ is acute $$\therefore$$ $$n = {1 \over 2}$$

$$\therefore$$ $$\overrightarrow v  = k\left( {{1 \over 2}\widehat i + {1 \over {\sqrt 2 }}\widehat j + {1 \over 2}\widehat k} \right),k \in R$$

$$\overrightarrow v .\,(\overrightarrow a  - \overrightarrow b ) = 0$$

$$(\sqrt 2  - a){1 \over 2} + ( - 1 - b){1 \over {\sqrt 2 }} + (1 - c){1 \over 2} = 0$$

$$ \Rightarrow {a \over 2} + {b \over {\sqrt 2 }} + {c \over 2} = {1 \over 2}$$

$$ \Rightarrow a + \sqrt 2 b + c = 1$$

Solutions · 1