Let → a = 3 ^ i + ^ j - ^ k and → c = 2 ^ i - 3 ^ j + 3 ^ k . If →...

Given,

\overset{\to}{a} = 3\overset{^}{i} + \overset{^}{j} - \overset{^}{k} and \overset{\to}{c} = 2\overset{^}{i} - 3\overset{^}{j} + 3\overset{^}{k} .

And $\overset{\to}{b}$ is a vector such that $\overset{\to}{a} = \overset{\to}{b} \times \overset{\to}{c}$ and $| \overset{\to}{b}{|}^{2} = 50,$

Now solving, $| \overset{\to}{a} | = | \overset{\to}{b} \times \overset{\to}{c} |$

$⇒ \sqrt{{3}^{2} + {1}^{2} + {1}^{2}} = | \overset{\to}{b} | \cdot \sqrt{{2}^{2} + {3}^{2} + {3}^{2}} \cdot sin\theta$

$⇒ \sqrt{11} = \sqrt{50} \cdot \sqrt{22} \cdot sin\theta$

$∴ sin\theta = \dfrac{1}{10}$ or $cos\theta = \dfrac{\sqrt{99}}{10}$

Now solving,\left|72 - {\left|\overset{\to}{b} + \overset{\to}{c}\right|}^{2}\right|$we get,</p><p>\left|72 - {\left|\overset{\to}{b} + \overset{\to}{c}\right|}^{2}\right|$

$= \left|72 - \left({\left|\overset{\to}{b}\right|}^{2} + {\left|\overset{\to}{c}\right|}^{2} + 2\left|\overset{\to}{b}\right|\left|\overset{\to}{c}\right|cos\theta\right)\right|$

$= \left|72 - \left(50 + 22 + 2 \times 5\sqrt{2} . \sqrt{22}\dfrac{\sqrt{99}}{10}\right)\right|$

= \left|72 - \left(72 + \dfrac{2 \times 5 \times 2 \times 11 \times 3}{10}\right\right|

$= \left|66\right| = 66$