A plane passing through the points (0,-1,0) and (0,0,1) and making an angle 4 with the pla…

Question

A plane passing through the points $(0,-1,0)$ and $(0,0,1)$ and making an angle $\dfrac\pi4$ with the plane $y-z+5=0$, also passes through the point:

Accepted Answer

Let ax + by + cz = 1 be the equation of the plane

it passed through point (0, --1, 0).

$$ \therefore $$ -b = 1

$$ \Rightarrow $$ b = -1

Also it passes through point (0, 0, 1)

$$ \therefore $$ c = 1

So the plane is ax - y + z = 1.

This plane an angle $${\pi  \over 4}$$ with the
plane y -- z + 5 = 0.

Normal to the plane ax - y + z = 1 is

$${\overrightarrow a }$$ = $$a\widehat i - \widehat j + \widehat k$$

Normal to the plane y -- z + 5 = 0 is

$${\overrightarrow b }$$ = $$\widehat j - \widehat k$$

cos $$\theta $$ = $$\left| {{{\overrightarrow a .\overrightarrow b } \over {\left| {\overrightarrow a } \right|\left| {\overrightarrow b } \right|}}} \right|$$

$$ \Rightarrow $$ $${1 \over {\sqrt 2 }}$$ = $$\left| {{{\overrightarrow a .\overrightarrow b } \over {\left| {\overrightarrow a } \right|\left| {\overrightarrow b } \right|}}} \right|$$

$$ \Rightarrow $$ $${{\left| {0 - 1 - 1} \right|} \over {\sqrt {{a^2} + 1 + 1} \sqrt {{1^2} + {1^2}} }}$$ = $${1 \over {\sqrt 2 }}$$

$$ \Rightarrow $$ ${a^2} + 2 = 4$

$$ \Rightarrow $$ a = $$ \pm $$ $$\sqrt 2 $$

$$ \therefore $$ Equation of plane

$$ \pm $$ $$\sqrt 2 $$x - y + z = 1

Now by checking each options you can see

equation - $$\sqrt 2 $$x - y + z = 1  satisfy by the point $$\left( {\sqrt 2 ,1,4} \right)$$.

Solutions · 1