If λ _ 1 < λ _ 2 are two values of λ such that the angle between...

Given:

{P}_{1}: \overset{\to}{r} \cdot \left(3\overset{^}{i} - 5\overset{^}{j} + \overset{^}{k}\right) = 7

So, normal vector is \overset{\to}{n}1 = 3\overset{^}{i} - 5\overset{^}{j} + \overset{^}{k}

{P}_{2}: \overset{\to}{r} \cdot \left(\lambda\overset{^}{i} + \overset{^}{j} - 3\overset{^}{k}\right) = 9

So, normal vector is \overset{\to}{n}2 = \lambda\overset{^}{i} + \overset{^}{j} - 3\overset{^}{k}

Now, angle between planes ${P}_{1} and {P}_{2}$ is

$\theta = \sin^{- 1}\left(\dfrac{2\sqrt{6}}{5}\right)$

$⇒ sin\theta = \dfrac{2\sqrt{6}}{5}$

$⇒ cos\theta = \dfrac{1}{5}$

Also,

$cos\theta = \dfrac{\overset{\to}{n}1 \cdot \overset{\to}{n}2}{\left|\overset{\to}{n}1\right|\left|\overset{\to}{n}2\right|}$

⇒ \dfrac{1}{5} = \dfrac{\left(3\overset{^}{i} - 5\overset{^}{j} + \overset{^}{k}\right) \cdot \left(\lambda\overset{^}{i} + \overset{^}{j} - 3\overset{^}{k}\right)}{\sqrt{35}\sqrt{{\lambda}^{2} + 10}}

$⇒ \dfrac{1}{5} = \left|\dfrac{3\lambda - 8}{\sqrt{35} \cdot \sqrt{{\lambda}^{2} + 10}}\right|$

Squaring both sides, we get

$⇒ \dfrac{1}{25} = \dfrac{9{\lambda}^{2} + 64 - 48\lambda}{35\left({\lambda}^{2} + 10\right)}$

$⇒ 7\left({\lambda}^{2} + 10\right) = 45{\lambda}^{2} + 320 - 240\lambda$

$⇒ 38{\lambda}^{2} + 250 - 310\lambda = 0$

$⇒ 19{\lambda}^{2} - 120\lambda + 125 = 0$

$⇒ 19{\lambda}^{2} - 95\lambda - 25\lambda + 125 = 0$

$⇒ x = 5, \dfrac{25}{19}$

Since, ${\lambda}_{2} > {\lambda}_{1}$, so ${\lambda}_{2} = 5, {\lambda}_{1} = \dfrac{25}{19}$

Perpendicular distance of point

\left(38{\lambda}_{1}, 10{\lambda}_{2}, 2\right) ≡ ( 50, 50, 2 )$from plane${P}_{1}$is</p><p>$= \left|\dfrac{3 \times 50 - 5 \times 50 + 2 - 7}{\sqrt{35}}\right| = \dfrac{105}{\sqrt{35}}$</p><p>Square of the distance is</p><p>$= \dfrac{105 \times 105}{35} = 315$