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Question

Mathematics Question on Vector Algebra

If λ(3i^+2j^6k^)\lambda\left(3\hat{i}+2\hat{j}-6\hat{k}\right) is a unit vector, then the values of λ\lambda are

A

±17\pm\frac{1}{7}

B

±7\pm7

C

±43\pm\sqrt{43}

D

±143\pm\frac{1}{\sqrt{43}}

Answer

±17\pm\frac{1}{7}

Explanation

Solution

Now, (3i^+2j^6k^)=32+22+(6)2|(3 \hat{ i }+2 \hat{ j }-6 \hat{ k })| =\sqrt{3^{2}+2^{2}+(-6)^{2}}
=9+4+36=\sqrt{9+4+36}
=49=7=\sqrt{49}=7
Since, λ(3i^+2j^6k^)\lambda(3 \hat{ i }+2 \hat{ j }-6 \hat{ k }) is a unit vector.
λ=±13i^+2j^6k^\therefore \lambda =\pm \frac{1}{|3 \hat{ i }+2 \hat{ j }-6 \hat{ k }|}
=±17=\pm \frac{1}{7}