Question

The reflection of the plane $ax+by+cz+d=0$ in the plane $a^{\prime }x+b^{\prime }y+c^{\prime }z+d^{\prime }=0$ is:

• A. $(a^{\prime 2}+b^{\prime 2}+c^{\prime 2})(a^{\prime }x+b^{\prime }y+c^{\prime }z+d^{\prime })=(a{a}^{\prime }+b{b}^{\prime }+c{c}^{\prime })(ax+by+cz+d)$
• B. $2(a^{\prime 2}+b^{\prime 2}+c^{\prime 2})(a^{\prime }x+b^{\prime }y+c^{\prime }z+d^{\prime })=(a{a}^{\prime }+b{b}^{\prime }+c{c}^{\prime })(ax+by+cz+d)$
• C. $(a{a}^{\prime }+b{b}^{\prime }+c{c}^{\prime })(a^{\prime }x+b^{\prime }y+c^{\prime }z+d^{\prime })=(a^{\prime 2}+b^{\prime 2}+c^{\prime 2})(ax+by+cz+d)$
• D. $2(a{a}^{\prime }+b{b}^{\prime }+c{c}^{\prime })(a^{\prime }x+b^{\prime }y+c^{\prime }z+d^{\prime })=(a^{\prime 2}+b^{\prime 2}+c^{\prime 2})(ax+by+cz+d)$

Answer 1

The correct option is D $2(a{a}^{\prime }+b{b}^{\prime }+c{c}^{\prime })(a^{\prime }x+b^{\prime }y+c^{\prime }z+d^{\prime })=(a^{\prime 2}+b^{\prime 2}+c^{\prime 2})(ax+by+cz+d)$

Given planes are $ax+by+cz+d=0\cdots \left(i\right)$
$a^{\prime }x+b^{\prime }y+c^{\prime }z+d^{\prime }=0\cdots \left(ii\right)$

Let $P(\alpha ,\beta ,\gamma )$ be an arbitrary point in the plane $\left(i\right)$ and $Q(p,q,r)$ be the reflection of the point $P$ in plane $\left(ii\right)$.
Locus of $Q$ will be the required reflection of plane $\left(i\right)$ in plane $\left(ii\right)$. Let $L$ be the mid point of $PQ$
Then $L(\frac{p+\alpha }{2},\frac{q+\beta }{2},\frac{r+\gamma }{2})$
$L$ lies in plane $\left(ii\right)$, so we get $a^{\prime }\left(\frac{p+\alpha }{2}\right)+b^{\prime }\left(\frac{q+\beta }{2}\right)+c^{\prime }\left(\frac{r+\gamma }{2}\right)+d^{\prime }=0$
$\Rightarrow a^{\prime }(p+\alpha )+b^{\prime }(q+\beta )+c^{\prime }(r+\gamma )+2d^{\prime }=0\cdots \left(iii\right)$
$D.R^{\prime }s$ of $PQ$ are $(\alpha -p,\beta -1,\gamma -r)$
Since $PQ$ is perpendicular to plane in equation $\left(iii\right)$, the $D.R^{\prime }s$ of $PQ$ and that of the normal of the plane are parallel. So, we get $\frac{\alpha -p}{a^{\prime }}=\frac{\beta -1}{b^{\prime }}=\frac{\gamma -r}{c^{\prime }}=k\left(\text{say}\right)$
$\therefore \alpha =p+a^{\prime }k,\beta =q+b^{\prime }k,\gamma =r+c^{\prime }k$
Putting the values of $\alpha ,\beta$ and $\gamma$ in equation $\left(iii\right)$, we get $a^{\prime }(2p+k{a}^{\prime })+b^{\prime }(2q+k{b}^{\prime })+c^{\prime }(2r+k{c}^{\prime })+2d^{\prime }=0$
$\Rightarrow 2(a^{\prime }p+b^{\prime }q+c^{\prime }r+d^{\prime })=-k(a^{\prime 2}+b^{\prime 2}+c^{\prime 2})\cdots \left(iv\right)$
Since $P(\alpha ,\beta ,\gamma )$ lies on plane $\left(i\right)$, we get $a\alpha +b\beta +c\gamma +d=0$
$\Rightarrow a(p+k{a}^{\prime })+b(q+k{b}^{\prime })+c(r+k{c}^{\prime })+d=0$
$\Rightarrow k=-\frac{ap+bq+cr+d}{a{a}^{\prime }+b{b}^{\prime }+c{c}^{\prime }}$
Now, putting the value of $k$ in equation $\left(iv\right)$, we get $2(a^{\prime }p+b^{\prime }q+c^{\prime }r+d^{\prime })=\frac{(a^{\prime 2}+b^{\prime 2}+c^{\prime 2})(ap+bq+cr+d)}{a{a}^{\prime }+b{b}^{\prime }+c{c}^{\prime }}$
$\therefore$ Locus of $Q(p,q,r)$ i.e equation of reflection of plane $\left(i\right)$ in plane $\left(ii\right)$ is, $2(a{a}^{\prime }+b{b}^{\prime }+c{c}^{\prime })(a^{\prime }x+b^{\prime }y+c^{\prime }z+d^{\prime })=(a^{\prime 2}+b^{\prime 2}+c^{\prime 2})(ax+by+cz+d)$