Question

Find the angle between vectors A=i-2j+k and vector B=4i-4j+7k.​

Answer 1

Answer:

θ = cos⁻¹ {19/(√6*9)}

Explanation:

Given

A = i - 2j + k

B = 4i - 4j + 7k

Let angle between A and B be θ.

|A| = |i - 2j + k|                            |           |B| = |4i - 4j + 7k

    = $\sqrt{1^{2}+(-2)^{2}+1^{2}$               |                = $\sqrt{4^{2}+(-4)^{2}+7^{2}$

    = √(1 + 4 +1)                          |                = √(16 + 16 + 49)

∴|A| = √6                                   |          ∴|B|  = √81 = 9

And

A.B = (i - 2j + k).(4i - 4j + 7k)

      = 4 + 8 + 7                 [∵i.i = j.j = k.k = 1 and rest is 0 in dot product]

⇒A.B = 19

We know that

cos(a,b) = a.b/(|a|*|b|)

∴cos (A,B) = A.B/(|A|*|B|)

⇒cos θ = 19/(√6*9)

∴θ = cos⁻¹ {19/(√6*9)}