If for two vector A- and B- sum A-B- is perpendicular to the difference A-B- The ratio of their magnitude is

The correct option is C 1 (A⃗+B⃗) is perpendicular to (A⃗ b⃗0) . Thus (A⃗+B⃗).(A⃗ B⃗)=0 or A^2+B⃗.A⃗ A⃗.B⃗ B^2=0 Because of commutativ ...

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Byju's Answer

Standard XII

Physics

Multiplication with Vectors

If for two ve...

Question

If for two vector $\to A$ and $\to B$ , sum $(\to A + \to B)$ is perpendicular to the difference $(\to A - \to B)$ . The ratio of their magnitude is

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Solution

The correct option is C $1$
$(\to A + \to B)$ is perpendicular to $(\to A - \to b 0)$ . Thus
$(\to A + \to B) . (\to A - \to B) = 0$
or $A^{2} + \to B . \to A - \to A . \to B - B^{2} = 0$
Because of commutative property of dot product $\to A . \to B = \to B . \to A$
$∴ A^{2} - B^{2} = 0$ or $A = B$
Thus the ratio of magnitude $A / B = 1$

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If for two vector →A and →B, sum (→A+→B) is perpendicular to the difference (→A−→B). The ratio of their magnitude is

The correct option is C 1(→A+→B) is perpendicular to (→A−→b0). Thus(→A+→B).(→A−→B)=0or A2+→B.→A−→A.→B−B2=0Because of commutative property of dot product →A.→B=→B.→A∴A2−B2=0 or A=BThus the ratio of magnitude A/B=1

If for two vector $\to A$ and $\to B$ , sum $(\to A + \to B)$ is perpendicular to the difference $(\to A - \to B)$ . The ratio of their magnitude is