Question

find \\(\vec{a} \times \vec{b}\\) and the angle \\(\theta\\) between \\(\vec{a}\\) and \\(\vec{b}\\), given that \\(\vec{a} = -13\hat{j} + 11\hat{k}\\) and \\(\vec{b} = -25\hat{i} + 12\hat{j}\\). \\(\vec{a} \times \vec{b} = \\) \\(\theta = \\) \\(^\circ\\)

Explanation:

Step1: Write vectors in component form

$\vec{A} = 0\hat{i} -13\hat{j} + 11\hat{k}$, $\vec{B} = -25\hat{i} + 12\hat{j} + 0\hat{k}$

Step2: Compute cross product via determinant

$$ \vec{A} \times \vec{B} =
LATEXBLOCK0
$$

$$ = \hat{i}[(-13)(0)-(11)(12)] - \hat{j}[(0)(0)-(11)(-25)] + \hat{k}[(0)(12)-(-13)(-25)] $$

$$ = -132\hat{i} - 275\hat{j} - 325\hat{k} $$

Step3: Calculate magnitudes of vectors

$|\vec{A}| = \sqrt{0^2 + (-13)^2 + 11^2} = \sqrt{169 + 121} = \sqrt{290}$
$|\vec{B}| = \sqrt{(-25)^2 + 12^2 + 0^2} = \sqrt{625 + 144} = \sqrt{769}$

Step4: Find magnitude of cross product

$|\vec{A} \times \vec{B}| = \sqrt{(-132)^2 + (-275)^2 + (-325)^2}$

$$ = \sqrt{17424 + 75625 + 105625} = \sqrt{198674} $$

Step5: Solve for angle $\theta$

Use $|\vec{A} \times \vec{B}| = |\vec{A}||\vec{B}|\sin\theta$, so:
$\sin\theta = \frac{|\vec{A} \times \vec{B}|}{|\vec{A}||\vec{B}|} = \frac{\sqrt{198674}}{\sqrt{290}\sqrt{769}}$
$\sin\theta = \frac{\sqrt{198674}}{\sqrt{290 \times 769}} = \frac{\sqrt{198674}}{\sqrt{223010}} \approx 0.949$
$\theta = \arcsin(0.949) \approx 71.6^\circ$

Answer:

$\vec{A} \times \vec{B} = -132\hat{i} - 275\hat{j} - 325\hat{k}$
$\theta \approx 72^\circ$ (rounded to nearest whole number, or $71.6^\circ$ for more precision)