Question

Assertion :$\pi :x+2y+3z=p$, $S:x^2+y^2+z^2-2x+4y-6z-2=0$

The plane $\pi$ cuts the sphere $S$ in a circle of maximum radius if $p=6$

Reason: The plane $\pi$ lies outside the sphere $S$ if $p=26$.

• A. Both Assertion and Reason are correct and Reason is the correct explanation for Assertion
• B. Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion
• C. Assertion is correct but Reason is incorrect
• D. Both Assertion and Reason are incorrect

Answer 1

The correct option is B Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion

$\pi :x+2y+3z=p$

$s:x^2+y^2+z^2-2x+4y-6z-2=0$

centre of sphere$=(1,-2,3)$

radius$=\sqrt{1^2+(-2{)}^2+3^2+2}=4$

$\therefore$ At the maximum circle i.e. when plane cuts sphere the perpendicular distance of the plne from centre should be zero.

$\therefore \frac{P-(1-4+9)}{\sqrt{1^2+2^2+3^2}}=0\Rightarrow p=6$

$\therefore$ assertion is true.

If $p=26$ then,

Perpendicular distance$=\frac{26-(1-4+9)}{\sqrt{1^2+2^2+3^2}}=\frac{10\sqrt{14}}{7}>4$

$\therefore$ reason is true.