Given two vectors A = 2î + 4ĵ + 5k̂ and B = î + 3ĵ + 6k̂, find the product A.B, and the magnitude of A and B. What is the angle between them?
(i) The product (A.B) is as follows:
A . B = (2î + 4ĵ + 5k̂) . (î + 3ĵ + 6k̂)
A . B = (2 × 1) + (4 × 3) + (5 × 6)
A . B = 2 + 12 + 30
A . B = 44 units.
(ii) The magnitude of A and B is as follows:
[tex]\sf{Magnitude\:of\:A = \sqrt{2^2 + 4^2 + 5^2}}[/tex]
[tex]\sf{|A| = \sqrt{4 + 16 + 25}}[/tex]
[tex]\sf{|A| = \sqrt{45}}[/tex]
[tex]\sf{|A| = 3\sqrt{5}}[/tex]
[tex]\sf{Magnitude\:of\:B = \sqrt{1^2 + 3^2 + 6^2}}[/tex]
[tex]\sf{|B| = \sqrt{1 + 9 + 36}}[/tex]
[tex]\sf{|B| = \sqrt{46}}[/tex]
(iii) Angle between the two vectors is as follows:
We know:
[tex]\sf{A . B = |A| |B| Cos\theta}[/tex]
[tex] = \sf{Cos \theta = \dfrac{A . B}{|A| |B|}}[/tex]
[tex] = \sf{Cos\theta = \dfrac{44}{3 \sqrt{5} \times \sqrt{46}}}[/tex]
[tex] = \sf{Cos\theta = \dfrac{44}{3 \sqrt{230}}}[/tex]
[tex] = \sf{\theta = Cos^{-1} \dfrac{44}{3\sqrt{230}}}[/tex]
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Question:-
Given two vectors A = 2î + 4ĵ + 5k̂ and B = î + 3ĵ + 6k̂, find the product A.B, and the magnitude of A and B. What is the angle between them?
Given:-
To find:-
Solution:-
(i) The product (A.B) is as follows:
A . B = (2î + 4ĵ + 5k̂) . (î + 3ĵ + 6k̂)
A . B = (2 × 1) + (4 × 3) + (5 × 6)
A . B = 2 + 12 + 30
A . B = 44 units.
(ii) The magnitude of A and B is as follows:
[tex]\sf{Magnitude\:of\:A = \sqrt{2^2 + 4^2 + 5^2}}[/tex]
[tex]\sf{|A| = \sqrt{4 + 16 + 25}}[/tex]
[tex]\sf{|A| = \sqrt{45}}[/tex]
[tex]\sf{|A| = 3\sqrt{5}}[/tex]
[tex]\sf{Magnitude\:of\:B = \sqrt{1^2 + 3^2 + 6^2}}[/tex]
[tex]\sf{|B| = \sqrt{1 + 9 + 36}}[/tex]
[tex]\sf{|B| = \sqrt{46}}[/tex]
(iii) Angle between the two vectors is as follows:
We know:
[tex]\sf{A . B = |A| |B| Cos\theta}[/tex]
[tex] = \sf{Cos \theta = \dfrac{A . B}{|A| |B|}}[/tex]
[tex] = \sf{Cos\theta = \dfrac{44}{3 \sqrt{5} \times \sqrt{46}}}[/tex]
[tex] = \sf{Cos\theta = \dfrac{44}{3 \sqrt{230}}}[/tex]
[tex] = \sf{\theta = Cos^{-1} \dfrac{44}{3\sqrt{230}}}[/tex]
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