Question

5. If Ā = 3i +6j – 2k , the direction of cosines of the
vector Ā are :-​

Answer 1

Answer:

I hope it would help you

Answer 2

Given,

We are given a  Ā = 3i +6j – 2k.

To find,

The direction of cosines.

Solution,

• First, we have to calculate the magnitude or resultant of Ā

• The resultant of the vector is calculated as $\sqrt{a^{2}+b^{2}+c^{2} }$.

• Similarly, the magnitude or resultant of   Ā = 3i +6j – 2k would be,

          $\sqrt{3^{2} +6^{2} +2^{2} }$ = $\sqrt{49}$ = 7.

• The direction cosine of any given vector can be determined by dividing the corresponding coordinate of a vector by the resultant.

• According to the question, the direction ratios are 3, 6, and -2 respectively.

Hence, The direction of cosines of Ā = 3i +6j – 2k. will be, 3/7, 6/7, and -2/7 (Option b)