Question

Answer this Question of iit jee . Chapter → Three dimensional Geometry​

Answer 1

EXPLANATION.

If the angle θ between the line.

⇒ (x + 1)/1 = (y - 1)/2 = (z - 2)/2.

The plane : 2x - y + √λz + 4 = 0.

⇒ sinθ = 1/3.

As we know that,

$\sf \implies cos \theta = \dfrac{a_{1}a_{2} + b_{1}b_{2} + c_{1} c_{2}}{\sqrt{a_{1}^{2} + b_{1}^{2} + c_{1}^{2} }\sqrt{a_{2}^{2} + b_{2}^{2} + c_{2}^{2} } }$

Put the values in the equation, we get.

⇒ a₁ = 1, b₁ = 2, c₁ = 2.

⇒ a₂ = 2, b₂ = - 1, c₂ = √λ.

$\sf \implies cos(90 - \theta) = \dfrac{(1)(2) + (2)(-1) + (2)(\sqrt{\lambda} )}{\sqrt{(1)^{2} + (2)^{2} + (2)^{2} }\sqrt{(2)^{2} + (-1)^{2} + (\sqrt{\lambda})^{2} } }$

$\sf \implies sin \theta = \dfrac{2 - 2 + 2\sqrt{\lambda} }{\sqrt{1 + 4 + 4} \sqrt{4 + 1 + \lambda} }$

$\sf \implies sin(\theta) = \dfrac{2\sqrt{\lambda} }{\sqrt{9}\sqrt{5 + \lambda} }$

$\sf \implies \dfrac{1}{3} = \dfrac{2\sqrt{\lambda} }{3 \sqrt{5 + \lambda} }$

$\sf \implies 1 = \dfrac{2 \sqrt{\lambda} }{\sqrt{5 + \lambda} }$

$\sf \implies \sqrt{5 + \lambda } = 2\sqrt{\lambda}$

Squaring on both sides of the equation, we get.

$\sf \implies (\sqrt{5 + \lambda} )^{2} = (2\sqrt{\lambda} )^{2}$

$\sf \implies 5 + \lambda = 4 \lambda$

$\sf \implies 5 = 3 \lambda$

$\sf \implies \lambda = \dfrac{5}{3}$

Option [A] is correct answer.

Answer 2

If the angle θ between the line.

⇒ (x + 1)/1 = (y - 1)/2 = (z - 2)/2.

The plane : 2x - y + √λz + 4 = 0.

⇒ sinθ = 1/3.

As we know that,

$\sf \implies cos \theta = \dfrac{a_{1}a_{2} + b_{1}b_{2} + c_{1} c_{2}}{\sqrt{a_{1}^{2} + b_{1}^{2} + c_{1}^{2} }\sqrt{a_{2}^{2} + b_{2}^{2} + c_{2}^{2} } }$

Put the values in the equation, we get.

⇒ a₁ = 1, b₁ = 2, c₁ = 2.

⇒ a₂ = 2, b₂ = - 1, c₂ = √λ.

$\sf \implies cos(90 - \theta) = \dfrac{(1)(2) + (2)(-1) + (2)(\sqrt{\lambda} )}{\sqrt{(1)^{2} + (2)^{2} + (2)^{2} }\sqrt{(2)^{2} + (-1)^{2} + (\sqrt{\lambda})^{2} } }$

$\sf \implies sin \theta = \dfrac{2 - 2 + 2\sqrt{\lambda} }{\sqrt{1 + 4 + 4} \sqrt{4 + 1 + \lambda} }$

$\sf \implies \dfrac{1}{3} = \dfrac{2\sqrt{\lambda} }{3 \sqrt{5 + \lambda} }$

Squaring on both sides of the equation, we get.

$\sf \implies (\sqrt{5 + \lambda} )^{2} = (2\sqrt{\lambda} )^{2}$

$\sf \implies 5 + \lambda = 4 \lambda$

$\sf \implies 5 = 3 \lambda$

$\sf \implies \lambda = \dfrac{5}{3}$

Option [A] is correct answer.