Question

Calculate the angle between the two vectors given by a=3i+3j+3k and b=2i+1j+3k

Answer 1

Question

Calculate the angle between the two vectors given by a=3i+3j+3k and b=2i+1j+3k .

Solution

Let the first vector be Vector A

Let the second vector be Vector B

According TO THe Question,  

$\vec{A} = 3 \hat{i} \ + 3\hat{j} \ + 3 \hat{k} \\\\\vec{B} = 2 \hat{i} \ + 1\hat{j} \ + 3 \hat{k}$

Now we know that according to the Cross Product Of Vectors ,

$\vec{A} . \vec{ B } = |A||B| \sin{ \phi } \\\\=( 3 \hat{i} \  + 3\hat{j} \ + 3 \hat{k} ).( 2 \hat{i} \  + 1\hat{j} \ + 3 \hat{k})\\\\3 \times 2 \times {1} + 3 \times 1 \times {1} + 3 \times 3 \times {1}\\\\= 18 \\\\| A | = Magnitude \: Of \: A = \sqrt{ 3^{2}+3^{2}+3^{2}   } = 3\sqrt{3} \\\\| B | = Magnitude \: Of \: B = \sqrt{ 2^{2}+1 ^{2}+3^{2}   } = \sqrt{14}  \\\\\\Substuting \ the \ required \ values \ -\\\\3\sqrt{42} \sin{ \phi }  = 18\\\\ \sin{ \phi } = \dfrac{\sqrt{42} }{7}$

$So, \\\phi = Sin \ Inverse \ Of ( \dfrac{\sqrt{42} }{7} )$