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

sin and cos angle between two vectors.

[tex](1) : \sf \displaystyle A = 2 \hat{i} + \hat{k} \ \ \ and \ \ \ B = - \hat{i} + 2 \hat{j} - 3 \hat{k}[/tex]

As we know that,

[tex]\sf \displaystyle a. b = |a| |b| cos \theta[/tex]

[tex]\sf \displaystyle | a \times b| = |a| |b| sin \theta[/tex]

Using this formula in this question, we get.

[tex]\sf \displaystyle |A| = \sqrt{(2)^{2} + (0)^{2} + (1)^{2} }[/tex]

[tex]\sf \displaystyle |A| = \sqrt{5}[/tex]

[tex]\sf \displaystyle |B| = \sqrt{(-1)^{2} + (2)^{2} + (-3)^{2} }[/tex]

[tex]\sf \displaystyle |B| = \sqrt{1 + 4 + 9}[/tex]

[tex]\sf \displaystyle |B| = \sqrt{14}[/tex]

[tex]\sf \displaystyle A \times B = \left|\begin{array}{ccc}\hat{i}&\hat{j}&\hat{k}\\2&0&1\\-1&2&-3\end{array}\right|[/tex]

[tex]\sf \displaystyle A \times B = \hat{i} [(0)(-3) - (2)(1)] - \hat{j} [(2)(-3) - (-1)(1)] + \hat{k} [(2)(2) - (-1)(0)][/tex]

[tex]\sf \displaystyle A \times B = - 2 \hat{i} - \hat{j} [- 6 + 1] + 4 \hat{k}[/tex]

[tex]\sf \displaystyle A \times B = - 2 \hat{i} + 5 \hat{j} + 4 \hat{k}[/tex]

[tex]\sf \displaystyle |A \times B| = \sqrt{(-2)^{2} + (5)^{2} + (4)^{2} }[/tex]

[tex]\sf \displaystyle |A \times B| = \sqrt{4 + 25 + 16}[/tex]

[tex]\sf \displaystyle |A \times B| = \sqrt{45}[/tex]

[tex]\sf \displaystyle a. b = |a| |b| cos \theta[/tex]

[tex]\sf \displaystyle (2 \hat{i} + \hat{k} ) . (- \hat{i} + 2 \hat{j} - 3 \hat{k} ) = \sqrt{5} \sqrt{14} \ cos \theta[/tex]

[tex]\sf \displaystyle - 2 -3 = \sqrt{70} \ cos \theta[/tex]

[tex]\sf \displaystyle - 5 = \sqrt{70} \ cos \theta[/tex]

[tex]\sf \displaystyle \boxed{cos \theta = \frac{-5}{\sqrt{70} }}[/tex]

[tex]\sf \displaystyle | a \times b| = |a| |b| sin \theta[/tex]

[tex]\sf \displaystyle \sqrt{45} = \sqrt{5} \sqrt{14} \ sin \theta[/tex]

[tex]\sf \displaystyle \sqrt{45} = \sqrt{70} \ sin \theta[/tex]

[tex]\sf \displaystyle sin \theta = \sqrt{\frac{45}{70} }[/tex]

[tex]\sf \displaystyle \boxed{sin \theta = \sqrt{\frac{9}{14} }}[/tex]

[tex]\sf \displaystyle (2) : A = 2 \hat{i} + 3 \hat{j} - 5 \hat{k} \ \ \ and \ \ \ B = - \hat{i} - \hat {j} - \hat{k}[/tex]

As we know that,

[tex]\sf \displaystyle a. b = |a| |b| cos \theta[/tex]

[tex]\sf \displaystyle | a \times b| = |a| |b| sin \theta[/tex]

Using this formula in this question, we get.

[tex]\sf \displaystyle |A| = \sqrt{(2)^{2} + (3)^{2} + (-5)^{2} }[/tex]

[tex]\sf \displaystyle |A| = \sqrt{4 + 9 + 25}[/tex]

[tex]\sf \displaystyle |A| = \sqrt{38}[/tex]

[tex]\sf \displaystyle |B| = \sqrt{(-1)^{2} + (-1)^{2} + (-1)^{2} }[/tex]

[tex]\sf \displaystyle |B| = \sqrt{1 + 1 + 1}[/tex]

[tex]\sf \displaystyle |B| = \sqrt{3}[/tex]

[tex]\sf \displaystyle A \times B = \left|\begin{array}{ccc}\hat{i}&\hat{j}&\hat{k}\\2&3&-5\\-1&-1&-1\end{array}\right|[/tex]

[tex]\sf \displaystyle A \times B = \hat{i} [(3)(-1) - (-1)(-5)] - \hat{j} [(2)(-1) - (-1)(-5)] + \hat{k} [(2)(-1) - (3)(-1)][/tex]

[tex]\sf \displaystyle A \times B = \hat{i} [- 3 - 5] - \hat{j} [- 2 - 5 ] + \hat{k} [- 2 + 3][/tex]

[tex]\sf \displaystyle A \times B = - 8 \hat{i} + 7 \hat{j} + \hat{k}[/tex]

[tex]\sf \displaystyle |A \times B| = \sqrt{(-8)^{2} + (7)^{2} + (1)^{2} }[/tex]

[tex]\sf \displaystyle |A \times B| = \sqrt{64 + 49 + 1}[/tex]

[tex]\sf \displaystyle |A \times B| = \sqrt{114}[/tex]

[tex]\sf \displaystyle a. b = |a| |b| cos \theta[/tex]

[tex]\sf \displaystyle (2 \hat{i} + 3 \hat{j} - 5 \hat{k} ) . (- \hat{i} - \hat{j} - \hat{k} ) = \sqrt{38} \sqrt{3} cos \theta[/tex]

[tex]\sf \displaystyle - 2 - 3 + 5 = \sqrt{114} \ cos \theta[/tex]

[tex]\sf \displaystyle \boxed{cos \theta = 0}[/tex]

[tex]\sf \displaystyle | a \times b| = |a| |b| sin \theta[/tex]

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

[tex]\sf \displaystyle \sqrt{114} = \sqrt{114} \ sin \theta[/tex]

[tex]\sf \displaystyle \boxed{sin \theta = 1}[/tex]