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Chapter 16: Problem 1502

\(\underline{\mathrm{x}}=(2,-6,3), \mathrm{y}=(1,2,-2)\) and \(\underline{\mathrm{x}}^{\wedge} \mathrm{y}=\theta\), then \(\sin \theta=\) (A) \([21 / \sqrt{(185)}]\) (B) \(-[\sqrt{(185) / 21]}\) (C) \(-[21 / \sqrt{(185)}]\) (D) \([\sqrt{(185) / 21]}\)

Short Answer

Expert verified
The short answer is: \(\sin{\theta} = \dfrac{\sqrt{249}}{21}\) (Option A).

Step by step solution

01

Calculate the cross product of the vectors

To calculate the cross product of the two vectors, \(\underline{x}\) and \(y\), we use the following formula: \[\underline{x}\times y=\begin{pmatrix} i & j & k \\ 2 & -6 & 3 \\ 1 & 2 & -2 \end{pmatrix}\] Using the determinants method, we find: \[\underline{x}\times y = (3 \times (-2) - 2 \times 2)i - (2 \times (-2) - 1 \times 3)j + (2 \times 2 - 1 \times (-6))k\] \[\underline{x}\times y = (-6 - 4)i - (-4 -3)j + (4 + 6)k = -10i + 7j + 10k\] So, \(\underline{x}\times y = (-10, 7, 10)\).
02

Calculate the magnitudes of the vectors

We need to calculate the magnitudes of the vectors \(\underline{x}\) and \(y\), denoted as \(|\underline{x}|\) and \(|y|\), respectively: \[|\underline{x}| = \sqrt{(2)^2 + (-6)^2 + (3)^2} = \sqrt{4 + 36 + 9} = \sqrt{49} = 7\] \[|y| = \sqrt{(1)^2 + (2)^2 + (-2)^2} = \sqrt{1 + 4 + 4} = \sqrt{9} = 3\]
03

Solve for \(\sin{\theta}\)

Now we have all the values needed to calculate \(\sin{\theta}\): \[\sin{\theta} = \dfrac{|\underline{x}\times y|}{|\underline{x}||y|} = \dfrac{\sqrt{(-10)^2 + 7^2 + 10^2}}{7\cdot 3} = \dfrac{\sqrt{100+49+100}}{21} = \dfrac{\sqrt{249}}{21}\]
04

Identify the correct option

Comparing our result with the given options, we can see that our answer matches option (A): \[\sin{\theta} = \dfrac{\sqrt{249}}{21}\] So the correct answer is option (A).

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Most popular questions from this chapter

If plane \(2 \mathrm{x}-2 \mathrm{y}+\mathrm{z}=-3\) express in form of \(x \cos \alpha+y \cos \beta+z \cos \gamma=p\), then perpendicular distance from origin to the plane is foot of perpendicular is and direction cosine is (A) \(1,[-(2 / 3),(2 / 3),-\overline{(1 / 3)}],-(2 / 3),(2 / 3),\\{(-1) / 3\\}\) (B) \(2,[-(2 / 3),(2 / 3),-(1 / 3)],-(2 / 3),(2 / 3),\\{(-1) / 3\\}\) (C) \(1,[(2 / 3),(2 / 3),-(1 / 3)],(2 / 3),(2 / 3),(1 / 3)\) (D) None of these

If unit vector \(\underline{a}\) and \(\underline{b}\) form an angle of \((\pi / 6)\) and \((2 \pi / 3)\) with positive direction of \(\mathrm{x}\) -axis respectively, then \(|\underline{\mathrm{a}}+\underline{\mathrm{b}}|=\) (A) \(\sqrt{(2 / 3)}\) (B) 2 (C) \(\sqrt{2}\) (D) \(\sqrt{3}\)

The foot of the perpendicular and perpendicular distance from point \((1,2,3)\) to plane \(x-2 y+2 z=5\) is and respectively (A) \([(11 / 9),(14 / 9),(31 / 9)],(2 / 3)\) (B) \([\\{(-11) / 9\\},\\{(-14) / 9\\},\\{(-31) / 9\\}],(2 / 3)\) (C) \([(11 / 9),(14 / 9),(31 / 9)],(2 / 3)\) (D) \([(11 / 9),(14 / 9),\\{(-31) / 9\\}],(2 / 3)\)

If angle between two units vectors \(\underline{a}\) and \(\underline{b}\) is \(\theta\), then \(\sin (\theta / 2)=\) (A) \(|\underline{a}+\underline{b}|\) (B) \((1 / 2)|\underline{a}-\underline{b}|\) (C) \(|\underline{a}-\underline{b}|\) (D) \((1 / 2)|\underline{a}+\underline{b}|\)

The angle between two unit vectors a and \(\underline{b}\) is \(\theta,|\underline{a}+\underline{b}|<1\) if (A) \(\theta=(\pi / 2)\) (B) \(\theta>(\pi / 3)\) (C) \((2 \pi / 3)<\theta<\pi\) (D) \(\theta=(\pi / 6)\)
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