find the sin and cos angle between the two vectors 1) A = 2i + k and B = -i + 2j - 3k2) A = 2i + 3j - 5k and B = -i - j - k​

Explanation:

[tex]\large\underline{\sf{Solution-1}}[/tex]

Given that,

[tex]\sf \:\vec{a} = 2\hat{i} + \hat{k} \\ \\ [/tex]

and

[tex]\sf \:\vec{b} = - \hat{i} + 2\hat{j} - 3 \hat{k} \\ \\ [/tex]

Now, Consider

[tex]\sf \: |\vec{a}| = |2\hat{i} + \hat{k}| = \sqrt{ {(2)}^{2} + {(1)}^{2} } = \sqrt{5} \\ \\ [/tex]

[tex]\sf \: |\vec{b}| = |- \hat{i} + 2\hat{j} - 3 \hat{k}| = \sqrt{ {( - 1)}^{2} + {(2)}^{2} + {( - 3)}^{2} } = \sqrt{14} \\ \\ [/tex]

[tex]\sf \:\vec{a}.\vec{b} = (2\hat{i} + \hat{k}).(- \hat{i} + 2\hat{j} - 3 \hat{k}) = - 2 - 3 = - 5 \\ \\ [/tex]

We know,

[tex]\sf \:cos \theta \: = \: \dfrac{\vec{a}.\vec{b}}{ |\vec{a}| \: |\vec{b}| } \\ \\ [/tex]

On substituting the values, we get

[tex]\sf \:cos \theta \: = \: \dfrac{ - 5}{ \sqrt{5} \times \sqrt{14} } \\ \\ [/tex]

[tex]\sf\implies \bf \:cos \theta \: = \: - \dfrac{ \sqrt{5} }{ \sqrt{14} } \\ \\ [/tex]

Now, Consider

[tex]\sf \:\vec{a} \times \vec{b} = \rm \begin{gathered}\sf \left | \begin{array}{ccc}\hat{i}&\hat{j}&\hat{k}\\2&0& 1\\ - 1&2& -3\end{array}\right | \end{gathered} = - 2 \hat{i} +5 \hat{j} +4 \hat{k}\\ \\ [/tex]

So,

[tex]\sf \: |\vec{a} \times \vec{b}| = | - 2\hat{i} +5 \hat{j} +4 \hat{k}| = \sqrt{4 + 25 + 16} = \sqrt{45} = 3 \sqrt{5} \\ \\ [/tex]

Now, We know

[tex]\sf \:sin\theta \: = \: \dfrac{ |\vec{a} \times \vec{b}| }{ |\vec{a}| \: |\vec{b}| } \\ \\ [/tex]

[tex]\sf \:sin\theta \: = \: \dfrac{3 \sqrt{5} }{ \sqrt{5} \times \sqrt{14} } \\ \\ [/tex]

[tex]\sf\implies \sf \:sin\theta \: = \: \dfrac{3}{ \sqrt{14} } \\ \\ [/tex]

[tex]\large\underline{\sf{Solution-2}}[/tex]

Given that,

[tex]\sf \:\vec{a} = 2\hat{i} + 3\hat{j} - 5 \hat{k} \\ \\ [/tex]

and

[tex]\sf \:\vec{b} = - \hat{i} - \hat{j} - \hat{k} \\ \\ [/tex]

Now, Consider

[tex]\sf \: |\vec{a}| = |2\hat{i} + 3\hat{j} - 5 \hat{k}| = \sqrt{4 + 9 + 25} = \sqrt{38} \\ \\ [/tex]

[tex]\sf \: |\vec{b}| = | - \hat{i} - \hat{j} - \hat{k}| = \sqrt{1 + 1 + 1} = \sqrt{3} \\ \\ [/tex]

[tex]\sf \:\vec{a}.\vec{b} = (2\hat{i} + 3\hat{j} - 5 \hat{k}).( - \hat{i} - \hat{j} - \hat{k}) = - 2 - 3 + 5 = 0 \\ \\ [/tex]

So,

[tex]\sf \:cos \theta \: = \: \dfrac{\vec{a}.\vec{b}}{ |\vec{a}| \: |\vec{b}| } \\ \\ [/tex]

[tex]\sf \:cos \theta \: = \: \dfrac{0}{ \sqrt{38} \times \sqrt{3} } \\ \\ [/tex]

[tex]\sf\implies \bf \:cos \theta \: = \: 0 \\ \\ [/tex]

Now,

[tex]\sf \:\vec{a} \times \vec{b} = \rm \begin{gathered}\sf \left | \begin{array}{ccc}\hat{i}&\hat{j}&\hat{k}\\2&3& - 5\\ - 1& - 1& -1\end{array}\right | \end{gathered} = - 8\hat{i} +7 \hat{j} + \hat{k}\\ \\ [/tex]

So,

[tex]\sf \: |\vec{a} \times \vec{b}| = | - 8\hat{i} +7\hat{j} +\hat{k}| = \sqrt{64 + 49 + 1} = \sqrt{114} \\ \\ [/tex]

So,

[tex]\sf \:sin\theta \: = \: \dfrac{ |\vec{a} \times \vec{b}| }{ |\vec{a}| \: |\vec{b}| } \\ \\ [/tex]

[tex]\sf \:sin\theta \: = \: \dfrac{ \sqrt{114} }{ \sqrt{38} \: \sqrt{3} } \\ \\ [/tex]

[tex]\sf \:sin\theta \: = \: \dfrac{ \sqrt{57} }{ \sqrt{19} \: \sqrt{3} } \\ \\ [/tex]

[tex]\bf\implies \: \bf \:sin\theta \: = \: 1 \\ \\ [/tex]