Question

Find the equation of the plane through the line
$P=ax+by+cz+d=0,$
$Q=a^{\prime }x+b^{\prime }y+c^{\prime }z+d^{\prime }=0$
and parallel to the line $\frac{x}{l}=\frac{y}{m}=\frac{z}{n}.$

• A. $P(a^{\prime }l+b^{\prime }m+c^{\prime }n)+Q(al+bm+cn)=0.$
• B. $P(a^{\prime }l+b^{\prime }m+c^{\prime }n)-Q(al+bm+cn)=0.$
• C. $Q(a^{\prime }l+b^{\prime }m+c^{\prime }n)-P(al+bm+cn)=0.$
• D. $Q(a^{\prime }l+b^{\prime }m+c^{\prime }n)+P(al+bm+cn)=0.$

Answer 1

Answer: (B)

$P(a^{\prime }l+b^{\prime }m+c^{\prime }n)-Q(al+bm+cn)=0.$

Given $P=ax+by+cz+d=0Q=a^{\prime }x+b^{\prime }y+c^{\prime }z+d^{\prime }=0$

Plane passing through above lines is

$P+\lambda Q=0$

$ax+by+cz+d+\lambda (a^{\prime }x+b^{\prime }y+c^{\prime }z+d^{\prime })=0\left(1\right)(a+a^{\prime }\lambda )x+(b+b^{\prime }\lambda )y+(c+c^{\prime }\lambda )z+d+d^{\prime }=0$

has normal line with d.r.'s $(a+a^{\prime }\lambda ,b+b^{\prime }\lambda ,c+c^{\prime }\lambda )$ and parallel to line $\frac{x}{l}=\frac{y}{m}=\frac{z}{n}\left(2\right)$

So normal line and $\left(2\right)$ are perpendicular

According to perpendicular conditions

Sum of product of d.r.'s is zero

$(a+a^{\prime }\lambda )l+(b+b^{\prime }\lambda )m+(c+c^{\prime }\lambda )n+d+d^{\prime }=0$

$\frac{-(al+bm+cn)}{a^{\prime }l+b^{\prime }m+c^{\prime }n}=\lambda$

Put $\lambda$ value in equation $\left(1\right)$

We get $P+\lambda Q=0$

$P+\left(\frac{-(al+bm+cn)}{a^{\prime }l+b^{\prime }m+c^{\prime }n}\right)Q=0P(a^{\prime }l+b^{\prime }m+c^{\prime }n)-Q(al+bm+cn)=0$