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The sine of the angle between the vectors î - 2 ĵ + 3 k̂ and 2 î + ĵ + k̂ is
- a)5/21
- b)5/√7
- c)3/√14
- d)5/2√7
Correct answer is option 'D'. Can you explain this answer?
Most Upvoted Answer
The sine of the angle between the vectors i - 2 j + 3 k and 2 i + j + ...
To find the angle between two vectors, we can use the dot product formula:
$\vec{a} \cdot \vec{b} = |\vec{a}| |\vec{b}| \cos \theta$
where $\theta$ is the angle between the two vectors. Rearranging for $\cos \theta$, we get:
$\cos \theta = \frac{\vec{a} \cdot \vec{b}}{|\vec{a}| |\vec{b}|}$
To find the sine of the angle, we can use the identity $\sin \theta = \sqrt{1 - \cos^2 \theta}$.
Now applying these formulas to the given vectors:
$\vec{a} = \begin{pmatrix} 1 \\ -2 \\ 3 \end{pmatrix}$ and $\vec{b} = \begin{pmatrix} 2 \\ 1 \\ 1 \end{pmatrix}$
$|\vec{a}| = \sqrt{1^2 + (-2)^2 + 3^2} = \sqrt{14}$
$|\vec{b}| = \sqrt{2^2 + 1^2 + 1^2} = \sqrt{6}$
$\vec{a} \cdot \vec{b} = (1)(2) + (-2)(1) + (3)(1) = 3$
Therefore,
$\cos \theta = \frac{\vec{a} \cdot \vec{b}}{|\vec{a}| |\vec{b}|} = \frac{3}{\sqrt{14}\sqrt{6}} = \frac{\sqrt{21}}{14}$
$\sin \theta = \sqrt{1 - \cos^2 \theta} = \sqrt{1 - \frac{21}{196}} = \frac{5}{14}$
Therefore, the answer is (b) $\frac{5}{14}$.
$\vec{a} \cdot \vec{b} = |\vec{a}| |\vec{b}| \cos \theta$
where $\theta$ is the angle between the two vectors. Rearranging for $\cos \theta$, we get:
$\cos \theta = \frac{\vec{a} \cdot \vec{b}}{|\vec{a}| |\vec{b}|}$
To find the sine of the angle, we can use the identity $\sin \theta = \sqrt{1 - \cos^2 \theta}$.
Now applying these formulas to the given vectors:
$\vec{a} = \begin{pmatrix} 1 \\ -2 \\ 3 \end{pmatrix}$ and $\vec{b} = \begin{pmatrix} 2 \\ 1 \\ 1 \end{pmatrix}$
$|\vec{a}| = \sqrt{1^2 + (-2)^2 + 3^2} = \sqrt{14}$
$|\vec{b}| = \sqrt{2^2 + 1^2 + 1^2} = \sqrt{6}$
$\vec{a} \cdot \vec{b} = (1)(2) + (-2)(1) + (3)(1) = 3$
Therefore,
$\cos \theta = \frac{\vec{a} \cdot \vec{b}}{|\vec{a}| |\vec{b}|} = \frac{3}{\sqrt{14}\sqrt{6}} = \frac{\sqrt{21}}{14}$
$\sin \theta = \sqrt{1 - \cos^2 \theta} = \sqrt{1 - \frac{21}{196}} = \frac{5}{14}$
Therefore, the answer is (b) $\frac{5}{14}$.
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Community Answer
The sine of the angle between the vectors i - 2 j + 3 k and 2 i + j + ...
To find the sine of the angle between two vectors, we can use the formula:
sinθ = |a x b| / |a||b|
where a and b are the two vectors and axb is their cross product.
First, let's find the cross product of the two vectors:
i - 2j + 3k x 2i + j + k
= (2(-2) - 3(1))i - (1(1) - 3(2))j + (1(1) - 2(2))k
= -7i - 5j - 3k
Next, let's find the magnitudes of the two vectors:
|a| = sqrt(1^2 + (-2)^2 + 3^2) = sqrt(14)
|b| = sqrt(2^2 + 1^2 + 1^2) = sqrt(6)
Finally, we can plug all these values into the formula for sine:
sinθ = |-7i - 5j - 3k| / (sqrt(14) * sqrt(6))
= sqrt(49 + 25 + 9) / (sqrt(14) * sqrt(6))
= sqrt(83) / (sqrt(14) * sqrt(6))
= (sqrt(83)/14) * (1/sqrt(6))
Simplifying the expression, we get:
sinθ = (sqrt(83)/14) * (sqrt(6)/6)
= sqrt(83/84)
Therefore, the sine of the angle between the two vectors is approximately 0.492, which is option (b) 5/
sinθ = |a x b| / |a||b|
where a and b are the two vectors and axb is their cross product.
First, let's find the cross product of the two vectors:
i - 2j + 3k x 2i + j + k
= (2(-2) - 3(1))i - (1(1) - 3(2))j + (1(1) - 2(2))k
= -7i - 5j - 3k
Next, let's find the magnitudes of the two vectors:
|a| = sqrt(1^2 + (-2)^2 + 3^2) = sqrt(14)
|b| = sqrt(2^2 + 1^2 + 1^2) = sqrt(6)
Finally, we can plug all these values into the formula for sine:
sinθ = |-7i - 5j - 3k| / (sqrt(14) * sqrt(6))
= sqrt(49 + 25 + 9) / (sqrt(14) * sqrt(6))
= sqrt(83) / (sqrt(14) * sqrt(6))
= (sqrt(83)/14) * (1/sqrt(6))
Simplifying the expression, we get:
sinθ = (sqrt(83)/14) * (sqrt(6)/6)
= sqrt(83/84)
Therefore, the sine of the angle between the two vectors is approximately 0.492, which is option (b) 5/
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The sine of the angle between the vectors i - 2 j + 3 k and 2 i + j + k isa)5/21b)5/√7c)3/√14d)5/2√7Correct answer is option 'D'. Can you explain this answer?
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