*This question has multiple parts. Therefore each part has been answered as a separate question on Clay6.com*

Solution :

Let the equation of the line passing through $(4,2,4)$ be : $a(x-4) +b(y-2) +C9z-4) =0$--------(1)

This plane is perpendicular to $2x+5y+4z+1=0$ and $4x+7y+6z+2=0$

$\therefore 2a+5b+4c=0$ --------(2)

$4a+7b+6c=0$ ----------(3)

Solving equation (2) and (3) we get,

$\Large\frac{a}{\begin{vmatrix}5 & 4 \\ 7 & 6 \end{vmatrix}}$=$\Large\frac{b}{\begin{vmatrix}4 & 2 \\ 6 & 4 \end{vmatrix}}$=$\Large\frac{c}{\begin{vmatrix}2 & 5 \\ 4 & 7 \end{vmatrix}}$$= \lambda$

(ie)$\large\frac{a}{30-28} =\frac{b}{16-12} =\frac{c}{14-20}$$=\lambda$

$a= 2 \lambda$

$b= 4 \lambda$

$c= -6 \lambda$

Substituting for a,b,c in equ (1) we get,

$2 \lambda(x-4)+4 \lambda(y-2)+(-6\lambda)(z-4) =0$

$2x+4y-6z+8=0$

=> $2x-8+4y-8-6z+24=0$

=> $x+2y-3z+4=0$

This is the required equation of the plane .

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