# The direction ratios of normal to the plane through the points (0,-1,0) and (0,0,1) and making an angle $\frac{\pi}{4}$  with the plane $y-z+5=0$ are :Option 1)  $\sqrt2,1,-1$Option 2)$2,-1,1$Option 3)  $2,\sqrt2,-\sqrt2$Option 4)  $2\sqrt3,1,-1$

Cartesian equation of plane passing through a given point and normal to a given vector -

$\left ( x-x_{0} \right )a+\left ( y-y_{0} \right )b+\left ( z-z_{0} \right )c= 0$

- wherein

$\vec{r}= x\hat{i}+y\hat{j}+z\hat{k}$

$\vec{a}= x_{0}\hat{i}+y_{0}\hat{j}+z_{0}\hat{k}$

$\vec{n}= a\hat{i}+b\hat{j}+c\hat{k}$

Putting in

$\left ( \vec{r}-\vec{a} \right )\cdot \vec{n}= 0$

We get $\left ( x-x_{0} \right )a+\left ( y-y_{0} \right )b+\left ( z-z_{0} \right )c= 0$

Angle between two planes (Cartesian form) -

Let the two planes be

$ax+by+cz+d=0\: and \: a_{1}x+b_{1}y+c_{1}z+d_{1}=0$

then the angle between them is defined as the angle between their normals

$\cos \Theta = \frac{a_{1}a_{2}+b_{1}b_{2}+c_{1}c_{2}}{\sqrt{a_{1}^{2}+b_{1}^{2}+c_{1}^{2}}\sqrt{a_{2}^{2}+b_{2}^{2}+c_{2}^{2}}}$

-

Equation of line passing through $\left ( 0,-1,0 \right )$ & $\left ( 0,0,1 \right )$  is   $\frac{x-0}{0}=\frac{y+1}{1}=\frac{z-0}{1}=\lambda$

Now, $\left | \frac{\bar{n_{1}}\cdot \cdot \bar{n_{2}}}{\left | n_{1} \right |\cdot \left | n_{2} \right |} \right |=\cos \frac{\pi }{4}$

$\left | \frac{-1-1}{\sqrt{2}\sqrt{a^{2}+2}} \right |=\frac{1}{\sqrt{2}}$

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

So direction ratios : $\left ( \sqrt{2},-1,1 \right )or\left ( -\sqrt{2} ,-1,1\right )$

So answer are (1) and (3)

Option 1)

$\sqrt2,1,-1$

Option 2)

$2,-1,1$

Option 3)

$2,\sqrt2,-\sqrt2$

Option 4)

$2\sqrt3,1,-1$

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