Question

The equation of the of the plane passing through the line $\frac{x-1}{5}=\frac{y+2}{6}=\frac{z-3}{4}$ and the point $(4,3,7)$ is

• A. $4x+8y+7z=41$
• B. $4x-8y+7z=41$
• C. $4x-8y-7z=41$
• D. $4x-8y+7z=39$

Answer 1

The correct option is A $4x+8y+7z=41$

Given that,

plane passes through line $\frac{x-1}{5}=\frac{y+2}{6}=\frac{2-3}{4}$ and point ( 4 , 3 , 7 )

Let the equation of line passing through (4 , 3 , 7 ) be a ( x- 4 ) + b ( y - 3) + c ( z - 7 ) = 0 ----eq.1

As the line passes through eq.1 , it should pass ( 1 , -2 , 3 ) and parallel to ( 5 , 6 , 4) .

$\Rightarrow a(1-4)+b(-2-3)+c(3-7)=0$

$\Rightarrow 3a+5b+4c=0$----eq.2

As the direction ratios ( 5 , 6 , 4 ) are parallel,

$\Rightarrow a.5+b.6+c.4=0$

$\Rightarrow 5a+6b+4c=0$----eq.3

Solving eq.2 and eq.3

$2a+b=0$ [ eq.3 - eq.2 ]

$b=-2a$ ----eq.4

Putting b in eq.2

$3a+5(-2a)+4c=0\Rightarrow -7a=-4c$

$\Rightarrow c=\frac{7}{4}a$----eq.5

From eq,4 and eq.5 in eq.1 , we get

$a(x-4)-2a(y-3)+\frac{7a}{4}(z-7)=0$

Dividing by 'a' on both sides,

$x-4-2y+6+\frac{7z}{4}-\frac{49}{4}=0$

On solving, $4x-8y+7z=41$ is the solution.