Question

Find the angle between the vectors A vector = 2 icap +4 jcap +6 kcap and B vector = 3 icap + j cap +2 kcap

THIS QUESTION IS 11 STANDARD QUESTION.
Please please please please help me.

Answer 1

$\vec{A} = 2\hat{i} + 4\hat{j} + 6\hat{k}$

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

''Now, For finding the angle, we usually use Dot product since it is easier than Cross product.''

$\vec{A}.\vec{B}  = (2\hat{i} + 4\hat{j} + 6\hat{k}).(3\hat{i} + \hat{j} + 2\hat{k})$

$\vec{A}.\vec{B} = 6 + 4 + 12$

$\vec{A}.\vec{B} = 22$

Now, for magnitude of Vector A and Vector B.

$|\vec{A}| = \sqrt{(2)^2 + (4)^2 + (6)^2}$

$|\vec{A}| = \sqrt{(4 + 16 + 36)}\\|\vec{A}| =  \sqrt{56}$

$|\vec{B}| = \sqrt{(3)^2 + (1)^2 + (2)^2}\\|\vec{B}| = \sqrt{9 + 1 + 4}\\|\vec{B}| = \sqrt{14}$

Now, Using the formula of Dot Product,

$\vec{A}. \vec{B} = |\vec{A}|| \vec{B}|Cos\alpha$

where α is the angle between Vector A and B.

∴ $22 = \sqrt{56} \sqrt{14} Cos\alpha$

∴ 22 = 2 × 14 × Cosα

∴ Cosα = 11/14

∴ α = Cos⁻¹(11/14)

Hope it helps.

Answer 2

Answer:

Let θ ( = 120°) be the angle between the two vectors P and Q.

|R|² = |P|² +  |Q|² + 2|P||Q|Cosθ

∴ |R|² = P² + Q² + 2PQ × -1/2

∴ R² = P² + Q² - PQ

Also, Let Resultant makes an angle α ( = 90) with smaller one.

Then,

Q = -PCosθ

∴ Q = -PCos120

∴ Q = -P × - 1/2

∴ Q = P/2

∴ Q = 40/2    [Since, P = 40].

∴ Q = 20 N.