Question

Assertion :The straight line $\frac{x-3}{-4}=\frac{y-4}{-7}=\frac{z+3}{13}$ lies in the plane $5x-y+z=8$ Reason: The straight line $\frac{x-x_1}{l}=\frac{y-y_1}{m}=\frac{z-z_1}{c}$ lies in the plane $ax+by+cz+d=0$ iff normal to the plane is perpendicular to the line & every point of the line satisfies the equation of the plane

• A. Both Assertion & Reason are individually true & Reason is correct explanation of Assertion
• B. Both Assertion & Reason are individually true but Reason is not the ,correct (proper) explanation of Assertion
• C. Assertion is true but Reason is false
• D. Assertion is false but Reason is true

Answer 1

The correct option is A Both Assertion & Reason are individually true & Reason is correct explanation of Assertion

Let $\frac{x-3}{-4}=\frac{y-4}{-7}=\frac{z+3}{13}=k$

Now $x=-4k+3$

$y=-7k+4$

$z=13k-3$

Putting these values in equation of plane we get

$5(-4k+3)+7k-4+13k-3=8$

Hence it satisfies

Also for line it's direction cosines $(\frac{-4}{r},\frac{-7}{r},\frac{13}{r})$

Normal$(5,-1,1)$

So $L_1\cdot N=-4\left(5\right)-1(-7)+13=-20+7+13=0$

Hence condition satisfies

Assertion and reason both correct and reason explains assertion