KCET · Maths · Vector Algebra
If \(|\mathbf{a}+\mathbf{b}|=|\mathbf{a}-\mathbf{b}|\), then
- A \(\mathbf{a}\) and \(\mathbf{b}\) are parallel.
- B \(\mathbf{a}\) and \(\mathbf{b}\) are coincident.
- C inclined to each other at \(60^{\circ}\).
- D \(\mathbf{a}\) and \(\mathbf{b}\) are perpendicular.
Answer & Solution
Correct Answer
(D) \(\mathbf{a}\) and \(\mathbf{b}\) are perpendicular.
Step-by-step Solution
Detailed explanation
Let \(\mathbf{a}\) and \(\mathbf{b}\) be two vectors \(|\mathbf{a}+\mathbf{b}|=|\mathbf{a}-\mathbf{b}|\)
On squaring on both sides, we get \(a^2+b^2+2 a b \cos \theta=a^2+b^2-2 a b \cos \theta\)
\(4 a b \cos \theta=0\)
If \(a=0, b=0\), then \(\cos \theta=0\)
\(\theta=\left(\frac{2 n-1}{2}\right) \pi\), where \(n \in N\)
Hence, \(\mathbf{a}\) and \(\mathbf{b}\) are perpendicular.
On squaring on both sides, we get \(a^2+b^2+2 a b \cos \theta=a^2+b^2-2 a b \cos \theta\)
\(4 a b \cos \theta=0\)
If \(a=0, b=0\), then \(\cos \theta=0\)
\(\theta=\left(\frac{2 n-1}{2}\right) \pi\), where \(n \in N\)
Hence, \(\mathbf{a}\) and \(\mathbf{b}\) are perpendicular.
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