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If `|vecA + vecB| = |vecA - vecB|`, then the angle between `vecA and vecB` will beA. `30^(@)`B. `45^(@)`C. `(60)^(@)`D. `90^(@)` |
Answer» Correct Answer - D Let `theta` be angle between the vectors `vecA and vecB`, Then `|vecA + vecB| = sqrt(A^(2) + B^(2) + 2AB cos theta)` `|vecA - vecB| = sqrt(A^(2) + B^(2) - 2AB cos theta)` According to given problem `|veA + vecB| = |vecA - vecB|` `therefore sqrt(A^(2) + B^(2) + 2AB cos theta) = sqrt(A^(2) + B^(2) - 2AB cos theta)` Squaring both sides, we get `A^(2) + B^(2) + 2AB cos theta = A^(2) + B^(2) - 2AB cos theta` `therefore 4AB cos theta = 0` As `A ne 0 , B ne 0 therefore cos theta = 0 or theta = 90^(@)` |
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