If for two vector vecA and vecB, sum ( vecA+ vecB) is perpendicular to the difference ( vecA- vecB). The ratio of their magnitude is

Question

If for two vector vecA and vecB, sum ( vecA+ vecB) is perpendicular to the difference ( vecA- vecB). The ratio of their magnitude is (A) 1 (B) 2 (C) 3 (D) None of these

Tardigrade Admin · Accepted Answer

Sol. (a) ( vecA+ vecB) is perpendicular to ( vecA- vecB). Thus( vecA+ vecB) ⋅( vecA- vecB)=0 or A2+ vecB ⋅ vecA- vecA ⋅ vecB-B2=0 Because of commutative property of dot product vecA ⋅ vecB= vecB ⋅ vecA ∴ A2-B2=0 or A=B Thus the ratio of magnitudes A / B=1

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

1

2

3

None of these

Sol. (a) $(A + B)$ is perpendicular to $(A - B)$ .
Thus $(A + B) \cdot (A - B) = 0$
or $A^{2} + B \cdot A - A \cdot B - B^{2} = 0$
Because of commutative property of dot product $A \cdot B = B \cdot A$
$∴ A^{2} - B^{2} = 0$ or $A = B$
Thus the ratio of magnitudes $A / B = 1$

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

1

2

3

None of these

Sol. (a) (A+B) is perpendicular to (A−B). Thus(A+B)⋅(A−B)=0 or 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 A/B=1

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

Sol. (a) $(A + B)$ is perpendicular to $(A - B)$ .
Thus $(A + B) \cdot (A - B) = 0$
or $A^{2} + B \cdot A - A \cdot B - B^{2} = 0$
Because of commutative property of dot product $A \cdot B = B \cdot A$
$∴ A^{2} - B^{2} = 0$ or $A = B$
Thus the ratio of magnitudes $A / B = 1$