Question

Find \(|\vec{a}+\vec{b}|\), if \(|\vec{a}|=3 \,and \,|\vec{b}|=4 \,and \,\vec{a}.\vec{b}=6\).

(a) 34

(b) \(\sqrt{37}\)

(c) 13

(d) \(\sqrt{23}\)

The question was asked in my homework.

This intriguing question originated from Product of Two Vectors-1 in chapter Vector Algebra of Mathematics – Class 12

Answer 1

The correct answer is (b) \(\sqrt{37}\)

The explanation is: \(|\vec{a}+\vec{b}|^2=(\vec{a}+\vec{b}).(\vec{a}+\vec{b})\)

=\(\vec{a}.\vec{a}+\vec{a}.\vec{b}+\vec{b}.\vec{a}+\vec{b}.\vec{b}\)

=\(|\vec{a}|^2+2(\vec{a}.\vec{b})+|\vec{b}|^2\)

=(3)^2+2(6)+(4)^2

=9+12+16=37

∴\(|\vec{a}+\vec{b}|=\sqrt{37}\)