MCQ
If $|\vec{a}|=2,|\vec{b}|=5$ and $|\vec{a} \times \vec{b}|=8$, then $|\vec{a} \cdot \vec{b}|$ is equal to :
  • A
    $5$
  • B
    $4$
  • $6$
  • D
    $3$

Answer

Correct option: C.
$6$
c
$|\vec{a}|=2,|\vec{b}|=5$

$|\vec{a} \times \vec{b}|=|\vec{a}||\vec{b}| \sin \theta=\pm 8$

$\sin \theta=\pm \frac{4}{5}$

$\therefore \vec{a} \cdot \vec{b}=|\vec{a}| \vec{b} \mid \cos \theta$

$=10 \cdot\left(\pm \frac{3}{5}\right)=\pm 6$

$|\vec{a} \cdot \vec{b}|=6$

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

Similar questions

A curve satisfying the initial condition $y(1)= 0$ satisfies the differential equation $x \frac{dy}{dx}= y -x^2$ the area bounded by the curve and the $x$ -axis is
The number of real roots of the equation $5 + |2^x - 1| = 2^x(2^x - 2)$ is
Let $g ( x )=\int_{0}^{ x } f( t ) dt ,$ where $f$ is continuous function in $[0,3]$ such that $\frac{1}{3} \leq f(t) \leq 1$ for all $t \in[0,1]$ and $0 \leq f( t ) \leq \frac{1}{2}$ for all $t \in(1,3]$ The largest possible interval in which $g (3)$ lies is :
If ${\tan ^{ - 1}}\frac{{a + x}}{a} + {\tan ^{ - 1}}\frac{{a - x}}{a} = \frac{\pi }{6}$,then ${x^2} =$
The area bounded by $y = x^2 + 2$ and $y = 2|x| -cos\,\pi x$ is equal to
Three dice are rolled. If the probability of getting different numbers on the three dice is $\frac{p}{q}$, where $p$ and $q$ are co-prime, then $q- p$ is equal to
The differential equation satisfied by the family of curves $y = ax\cos \,\left( {\frac{1}{x} + b} \right)$, where $a, b$ are parameters, is
The order of the differential equation whose general solution is given by $y = ({c_1} + {c_2})$ $\cos (x + {c_3}) - {c_4}{e^{x + {c_5}}},$ where ${c_1},\;{c_2},\;{c_3},\;{c_4},\;{c_5}$ are arbitrary constants, is
The sum of the series $aC_0 + (a + b)C_1 + (a + 2b)C_2 + ..... + (a + nb)C_n$ is where $Cr's$ denotes combinatorial coefficient in the expansion of $(1 + x)^n, n \in N$
If  $\int {\frac{{dx}}{{{{\cos }^3}\,x\sqrt {2\,\sin \,2x} }} = {{(\tan \,\,x)}^A} + C{{(\tan \,\,x)}^B} + k,} $ where $k$ is a constant of integration, then $A+ B + C$ equals