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Question: The unit vector parallel to the resultant of the vectors \(\overset{\rightarrow}{A} = 4\widehat{i} ...

The unit vector parallel to the resultant of the vectors

A=4i^+3j^+6k^\overset{\rightarrow}{A} = 4\widehat{i} + 3\widehat{j} + 6\widehat{k}andB=i^+3j^8k^\overset{\rightarrow}{B} = - \widehat{i} + 3\widehat{j} - 8\widehat{k} is –

A

(a)17(3i^+6j^2k^)\frac{1}{7}(3\widehat{i} + 6\widehat{j} - 2\widehat{k})

A

(b)17(3i^+6j^+2k^)\frac{1}{7}(3\widehat{i} + 6\widehat{j} + 2\widehat{k})

A

(c)149(3i^+6j^+2k^)\frac{1}{49}(3\widehat{i} + 6\widehat{j} + 2\widehat{k})

A

(d)149(3i^+6j^2k^)\frac{1}{49}(3\widehat{i} + 6\widehat{j} - 2\widehat{k})

Explanation

Solution

(a)

C=A+B\overset{\rightarrow}{C} = \overset{\rightarrow}{A} + \overset{\rightarrow}{B} = (3i^+6j^2k^)(3\widehat{i} + 6\widehat{j} - 2\widehat{k})

C^=CC\widehat{C} = \frac{\overset{\rightarrow}{C}}{|C|} = 3i^+6j^2k^32+62+(2)2\frac{3\widehat{i} + 6\widehat{j} - 2\widehat{k}}{\sqrt{3^{2} + 6^{2} + (–2)^{2}}} = 3i^+6j^2k^7\frac{3\widehat{i} + 6\widehat{j} - 2\widehat{k}}{7}