If |a-b|=|a+b| then the angle between a and b will be - - Vidyadip

MCQ

If $|\vec{a}-\vec{b}|=|\vec{a}+\vec{b}|$ then the angle between $\vec{a}$ and $\vec{b}$ will be -

A
$90^{\circ}$
B
$45^{\circ}$
C
$30^{\circ}$
D
0

✓

Answer

self

Need a full question paper?

Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.

Start Generating Free

Explore more

More "M.C.Q (1 Marks)" from [ Question Bank ]→[ Question Bank ] — all question types→Maths STD 12 Science question papers→Gujarat Board STD 12 Science question papers→Gujarat Board question paper generator→

Similar questions

If $f(x) = x^2 + 2bx + 2c^2$ and $g(x) = -x^2 -2cx + b^2$ such that $\min . f(x) > \max . g(x),$ then relation between $b$ and $c$ is

The sum $\sum\limits_{n = 1}^\infty {{{\cot }^{ - 1}}} \left( {\frac{{2\left( {\sum\limits_{k = 1}^n k } \right) - 1}}{3}} \right)$ is equal to

$f(x) = \left| {\begin{array}{*{20}{c}}
{{x^3}}&{{x^2}}&{3{x^2}}\\
1&{ - 6}&4\\
p&{{p^2}}&{{p^3}}
\end{array}} \right|$ , here $ p $ is a constant, then ${{{d^3}f(x)} \over {d{x^3}}}$ is

$\int_0^\pi {\frac{{dx}}{{1 + \sin x}}} = $

$\int_{}^{} {\frac{{{{({{\tan }^{ - 1}}x)}^3}}}{{1 + {x^2}}}\,dx = } $

The area enclosed by the closed curve $C$ given by the differential equation $\frac{d y}{d x}+\frac{x+a}{y-2}=0, y(1)=0$ is $4 \pi$.

Let $P$ and $Q$ be the points of intersection of the curve $C$ and the $y$-axis. If normals at $P$ and $Q$ on the curve $C$ intersect $x$-axis at points $R$ and $S$ respectively, then the length of the line segment $RS$ is

If $f\left( x \right) = {\sin ^{ - 1}}\left( {\frac{{2 \times {3^x}}}{{1 + {9^x}}}} \right)$, then $f'(-\frac {1}{2})$ equals

If $\sin\text{y}=\text{x}\cos(\text{a}+\text{y}),$ then $\frac{\text{dy}}{\text{dx}}$ is equal to:

$\frac{\cos^2(\text{a}+\text{y})}{\cos\text{a}}$
$\frac{\cos\text{a}}{\cos^2(\text{a}+\text{y})}$
$\frac{\sin^2\text{y}}{\cos\text{a}}$
$\text{None of these.}$

A quadratic polynomial $P(x)$ satisfies the conditions, $P(0) = P(1) = 0\, \&\,\int\limits_0^1 {} P(x) dx = 1$. The leading coefficient of the quadratic polynomial is :

An integrating factor for the differential equation $(1 + {y^2})dx - ({\tan ^{ - 1}}y - x)dy = 0$