Question

Calculus 3 Review Planes

Answer 1

Step-by-step explanation:

Given planes in vector form is given by

$\vec{r}.( 3\hat{i} - 2\hat{j} + 5\hat{k}) = - 8 \\ \vec{r}.( 5\hat{i} - \hat{j} + 3\hat{k}) = 9$

Here,

$\vec{n_{1}} = 3\hat{i} - 2\hat{j} + 5\hat{k} \\ \vec{n_{2}} = 5\hat{i} - \hat{j} + 3\hat{k}$

So, acute angle between them is given by

$\cos( \theta) = \bigg | \frac{ \vec{n_{1} }.\vec{n_{2} }}{ |\vec{n_{1} }| . |\vec{n_{2} }| } \bigg|$

$\implies \cos( \theta) = \bigg | \frac{ (3 \hat{i} - 2 \hat{j} + 5 \hat{k}).(5 \hat{i} - \hat{j} + 3 \hat{k})}{ |3 \hat{i} - 2 \hat{j} + 5 \hat{k}| . |5 \hat{i} - \hat{j} + 3 \hat{k}| } \bigg|$

$\implies \cos( \theta) = \bigg | \frac{(15 + 2 + 15)}{ \sqrt{9 + 4 + 25} . \sqrt{25 + 1 + 9 }} \bigg|$

$\implies \cos( \theta) = \bigg | \frac{ 32 }{ \sqrt{38} . \sqrt{35}} \bigg|$

$\implies \cos( \theta) = \bigg | \frac{ 32}{ \sqrt{1330} } \bigg|$

$\implies \theta= \cos^{ - 1} \bigg( \frac{ 32}{ \sqrt{1330} } \bigg)$