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

A⃗+B⃗ is perpendicular to A⃗ B⃗. Thus, (A⃗+B⃗)· (A⃗ B⃗)=0 A^2+B⃗·A⃗ A⃗·B⃗ B^2=0 Because of commutative property of dot product A⃗·B⃗=B⃗·A⃗ ∴ A^2 B^2=0 or A ...

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Standard VII

History

Maharani Tarabai

If for two ve...

Question

If for two vectors $\to A$ 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 A $1$
$\to A + \to B$ is perpendicular to $\to A - \to B$ .

Thus,
$(\to A + \to B) \cdot (\to A - \to B) = 0$

$A^{2} + \to B \cdot \to A - \to A \cdot \to B - B^{2} = 0$

Because of commutative property of dot product

$\to A \cdot \to B = \to B \cdot \to A$

$∴ A^{2} - B^{2} = 0$ or $A = B$

Thus the ratio of magnitudes, $\frac{A}{B} = 1$ .

Final answer: (a).

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Maharani Tarabai

Standard VII History

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

The correct option is A 1→A+→B is perpendicular to →A−→B. Thus, (→A+→B)⋅(→A−→B)=0 A2+→B⋅→A−→A⋅→B−B2=0 Because of commutative property of dot product →A⋅→B=→B⋅→A ∴ A2−B2=0 or A=B Thus the ratio of magnitudes, AB=1. Final answer: (a).

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