3D Geometry 3 Question 21
####21. The direction ratios of normal to the plane through the points $(0,-1,0)$ and $(0,0,1)$ and making an angle $\pi / 4$ with the plane $y-z+5=0$ are
(2019 Main, 11 Jan I)
(a) $2,-1,1$
(b) $\sqrt{2}, 1,-1$
(c) $2, \sqrt{2},-\sqrt{2}$
(d) $2 \sqrt{3}, 1,-1$
Show Answer
Answer:
(b,c)
Solution:
- Let the equation of plane be
$ a(x-0)+b(y+1)+c(z-0)=0 $
$[\because$ Equation of plane passing through a point $\left(x_{1}, y_{1}, z_{1}\right)$ is given by $\left.a\left(x-x_{1}\right)+b\left(y-y_{1}\right)+c\left(z-z_{1}\right)=0\right]$ $\Rightarrow a x+b y+c z+b=0$
Since, it also passes through $(0,0,1)$, therefore, we get
$ c+b=0 $
Now, as angle between the planes
$ a x+b y+c z+b=0 $
and
$ y-z+5=0 \text { is } \frac{\pi}{4} $
$\therefore \cos \frac{\pi}{4} = \frac {\left|\mathbf{n} _ {1} \cdot \mathbf{n}_{2}\right|}{\left|\mathbf{n} _ {1}\right|\left|\mathbf {n} _{2}\right|} ;$ where $\mathbf{n} _ {1} = a \hat {\mathbf{i}} + b \hat {\mathbf{j}} + c \hat {\mathbf{k}} $
and $\quad \mathbf{n}_{2}=0 \hat{\mathbf{i}}+\hat{\mathbf{j}}-\hat{\mathbf{k}}$
$ \begin{aligned} \Rightarrow \quad \frac{1}{\sqrt{2}} & =\frac{|(a \hat{\mathbf{i}}+b \hat{\mathbf{j}}+c \hat{\mathbf{k}}) \cdot(0 \hat{\mathbf{i}}+\hat{\mathbf{j}}-\hat{\mathbf{k}})|}{\sqrt{a^{2}+b^{2}+c^{2}} \sqrt{0+1+1}} \\ & =\frac{|b-c|}{\sqrt{a^{2}+b^{2}+c^{2}} \sqrt{2}} \end{aligned} $
$\Rightarrow \quad a^{2}+b^{2}+c^{2}=|b-c|^{2}=(b-c)^{2}=b^{2}+c^{2}-2 b c$
$\Rightarrow \quad a^{2}=-2 b c$
$\Rightarrow \quad a^{2}=2 b^{2}$
$\Rightarrow \quad a= \pm \sqrt{2} b$
[Using Eq. (ii)]
$\Rightarrow$ Direction ratios $(a, b, c)=( \pm \sqrt{2}, 1,-1)$
So, options (b) and (c) are correct because
$2, \sqrt{2},-\sqrt{2}$ and $\sqrt{2}, 1,-1$. are multiple of each other.
