Question

sine of angle between the vectors 3 ICAP + 2 J cap + 2 k cap and two ICAP - 2 J cap + 4 k cap is​

Answer 1

Answer:

$sin\theta=\sqrt{\frac{77}{102}}$

Explanation:

Formula used:

$\text{The angle between the the vectors }\vec{a}\text{ and }\vec{b}\text{ is}$

$\boxed{sin\theta=\frac{|\vec{a}\times\vec{b}|}{|\vec{a}||\vec{b}|}}$

$\vec{a}=3\vec{i}+2\vec{j}+2\vec{k}$

$\vec{b}=2\vec{i}-2\vec{j}+4\vec{k}$

$\vec{a}\times\vec{b}=\left|\begin{array}{ccc}\vec{i}&\vec{j}&\vec{k}\\3&2&2\\2&-2&4}\end{array}\right|$

$\vec{a}\times\vec{b}=\vec{i}(8+4)-\vec{j}(12-4)+\vec{k}(-6-4)$

$\implies\:\vec{a}\times\vec{b}=12\vec{i}-8\vec{j}-10\vec{k}$

$\implies\:\vec{a}\times\vec{b}=2(6\vec{i}-4\vec{j}-5\vec{k})$

$|\vec{a}\times\vec{b}|=2\sqrt{36+25+16}$

$|\vec{a}\times\vec{b}|=2\sqrt{77}$

$|\vec{a}|=\sqrt{9+4+4}=\sqrt{17}$

$|\vec{b}|=\sqrt{4+4+16}=\sqrt{24}=2\sqrt{6}$

Now,

$sin\theta=\frac{|\vec{a}\times\vec{b}|}{|\vec{a}|\:|\vec{b}|}$

$\implies\:sin\theta=\frac{2\sqrt{77}}{\sqrt{17}\:2\sqrt{6}}$

$\implies\:sin\theta=\frac{\sqrt{77}}{\sqrt{17}\sqrt{6}}$

$\implies\:sin\theta=\frac{\sqrt{77}}{\sqrt{102}}$

$\implies\:\boxed{sin\theta=\sqrt{\frac{77}{102}}}$

Answer 2

Angle, $sin\theta=\dfrac{12i-10j-8k}{18.87}$

Explanation:

It is given that,

$\vec{A}=3i+2j+2k$

$\vec{B}=i-2j+4k$

If we want to find the sine of angle between two vectors, it can be calculated using cross product formula as :

$A\times B=|A||B| sin\theta$

Magnitude of vector A, $|A|=\sqrt{3^2+2^2+2^2}=4.12$

Magnitude of vector B, $|B|=\sqrt{1^2+2^2+4^2}=4.582$

$A\times B=(3i+2j+2k)\times (i-2j+4k)$

$A\times B=\begin{pmatrix}12&-10&-8\end{pmatrix}$

$A\times B=12i-10j-8k$

$sin\theta=\dfrac{A\times B}{|A||B|}$

$sin\theta=\dfrac{12i-10j-8k}{4.12\times 4.582}$

$sin\theta=\dfrac{12i-10j-8k}{18.87}$

So, the sine of angle between A and B is $\dfrac{12i-10j-8k}{18.87}$. Hence, this is the required solution.

Learn more,

Cross product

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