If |a + b| |a - b|,then the angle between a | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) Given the inequality $|a + b| > |a - b|$.Squaring both sides,we get:$|a + b|^2 > |a - b|^2$$(a + b) \cdot (a + b) > (a - b) \cdot (a - b)$$|a|^2 +

Vedclass · Accepted Answer

Acute

Vedclass · Answer

(A) Given the inequality $|a + b| > |a - b|$.Squaring both sides,we get:$|a + b|^2 > |a - b|^2$$(a + b) \cdot (a + b) > (a - b) \cdot (a - b)$$|a|^2 + |b|^2 + 2(a \cdot b) > |a|^2 + |b|^2 - 2(a \cdot b)$$2(a \cdot b) > -2(a \cdot b)$$4(a \cdot b) > 0$$a \cdot b > 0$Since $a \cdot b = |a||b| \cos 	heta$,we have $|a||b| \cos 	heta > 0$.Since the magnitudes $|a|$ and $|b|$ are positive,$\cos 	heta > 0$.This implies that the angle $	heta$ between $a$ and $b$ must be acute,i.e.,$0 \le 	heta < \frac{\pi}{2}$.

If $|a + b| > |a - b|$,then the angle between $a$ and $b$ is

Similar Questions

The locus of a point equidistant from two given points $a$ and $b$ is given by

Let $v_1, v_2, v_3, v_4$ be unit vectors in the $XY$-plane,one each in the interior of the four quadrants. Which of the following statements is necessarily true?

If $a = 4 \hat{i} + 6 \hat{j}$,$b = 3 \hat{j} + 4 \hat{k}$,and $c$ is the projection vector of $a$ on $b$,then $c$ and $|c|$ respectively are

If $a, b, c$ are non-coplanar vectors,then the point of intersection of the line passing through the points $2a+3b-c$ and $3a+4b-2c$ with the line joining the points $a-2b+3c$ and $a-6b+6c$ is

If $a$ is a vector of magnitude $7$ and $b$ is a vector of magnitude $8$,then the maximum value of $|a \cdot b|$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module