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Find the angle between the vectors ????=????+???? and ????=−????+????. (Give an exact answer. Use symbolic notation and fractions where needed.)

Sagot :

Answer:

The angle between them is 60 degrees

Step-by-step explanation:

Given

[tex]a = 2i + j -3k[/tex]

[tex]b = 3i - 2j -k[/tex]

Required

The angle between them

The cosine of the angle between them is:

[tex]\cos(\theta) = \frac{a\cdot b}{|a|\cdot |b|}[/tex]

First, calculate a.b

[tex]a \cdot b =(2i + j -3k) \cdot (3i - 2j -k)[/tex]

Multiply the coefficients of like terms

[tex]a \cdot b =2 * 3 - 1 * 2 - 3 * -1[/tex]

[tex]a \cdot b =7[/tex]

Next, calculate |a| and |b|

[tex]|a| = \sqrt{2^2 + 1^2 + (-3)^2[/tex]

[tex]|a| = \sqrt{14[/tex]

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

[tex]|b| = \sqrt{14}[/tex]

Recall that:

[tex]\cos(\theta) = \frac{a\cdot b}{|a|\cdot |b|}[/tex]

This gives:

[tex]\cos(\theta) = \frac{7}{\sqrt{14} * \sqrt{14}}[/tex]

[tex]\cos(\theta) = \frac{7}{14}[/tex]

[tex]\cos(\theta) = 0.5[/tex]

Take arccos of both sides

[tex]\theta =\cos^{-1}(0.5)[/tex]

[tex]\theta =60^o[/tex]