If a + b + c =0,| a |=3, | b |=5,| c |=7,then the angle between a… - Vidyadip

MCQ

If $\vec{\text{a}}+\vec{\text{b}}+\vec{\text{c}}=\vec{0},|\vec{\text{a}}|=3,\big|\vec{\text{b}}\big|=5,|\vec{\text{c}}|=7,$then the angle between $\vec{\text{a}}$ and $\vec{\text{b}}$ is :

A
$\frac{\pi}{6}$
B
$\frac{2\pi}{3}$
C
$\frac{5\pi}{3}$
✓
$\frac{\pi}{3}$

✓

Answer

Correct option: D.

$\frac{\pi}{3}$

Given, $|\vec{\text{a}}|=3,\big|\vec{\text{b}}\big|=5$ and $|\vec{\text{c}}|=7\dots(1)$
Let $\theta$ be the angle between $\vec{\text{a}}$ and $\vec{\text{b}}.$
Given that
$\vec{\text{a}}+\vec{\text{b}}+\vec{\text{c}}=0$
$\Rightarrow\vec{\text{a}}+\vec{\text{b}}=-\vec{\text{c}}$
$\Rightarrow\big|\vec{\text{a}}+\vec{\text{b}}\big|=|-\vec{\text{c}}|^2$
$\Rightarrow|\vec{\text{a}}|^2+\big|\vec{\text{b}}\big|^2+2\vec{\text{a}}.\vec{\text{b}}=|\vec{\text{c}}|^2$
$\Rightarrow2\vec{\text{a}}.\vec{\text{b}}=|\vec{\text{c}}|^2-|\vec{\text{a}}|^2-\big|\vec{\text{b}}\big|^2$
$\Rightarrow2\vec{\text{a}}.\vec{\text{b}}=7^2-3^3-5^2\ [$using $(1)]$
$\Rightarrow2\vec{\text{a}}.\vec{\text{b}}=15$
$\Rightarrow2|\vec{\text{a}}|\big|\vec{\text{b}}\big|\cos\theta=15$
$\Rightarrow2(3)(5)\cos\theta=15\ [$using $(1)]$
$\Rightarrow\cos\theta=\frac{1}{2}$
$\therefore\theta=\frac{\pi}{3}$

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 Vector Algebra→Vector Algebra — all question types→Maths STD 12 Science question papers→Gujarat Board STD 12 Science question papers→Gujarat Board question paper generator→

Similar questions

Let $f\left( x \right) = x\sqrt {x\sqrt {x\sqrt {x.......\infty } } } \left( {x > 0} \right)$ ,then $f'(3)$ is equal to

$\text{f(x)}=\begin{cases}\frac{\sqrt{1+\text{px}}-\sqrt{1-\text{px}}}{\text{x}},&\text{if }0\leq\text{x}<0\\\frac{2\text{x}+1}{\text{x}-2},&\text{if }0\leq\text{x}\leq1\end{cases}$ is continuous in the interval $[-1, 1],$ then $p$ is equal to :

If $a + b + c = 0$, then the solution of the equation $\left| {\,\begin{array}{*{20}{c}}{a - x}&c&b\\c&{b - x}&a\\b&a&{c - x}\end{array}\,} \right| = 0$ is

Choose the correct answer from the given four options.The solution of the equation $(2\text{y}-1)\text{dx}-(2\text{x}+3)\text{dy}=0$ is:

If $y = y ( x )$ is the solution of the differential equation $\left(1+ e ^{2 x }\right) \frac{ dy }{ dx }+2\left(1+ y ^{2}\right) e ^{ x }=0$ and $y (0)=0$, then $6\left( y ^{\prime}(0)+\left( y \left(\log _{ e } \sqrt{3}\right)\right)^{2}\right)$ is equal to

If $\omega $ is cube root of unity, then root of the equation $\left| {\begin{array}{*{20}{c}}
{x + 2}&\omega &{{\omega ^2}} \\
\omega &{x + 1 + {\omega ^2}}&1 \\
{{\omega ^2}}&1&{x + 1 + \omega }
\end{array}} \right| = 0$ is

The vector in the direction of the vector $\hat{i}-2 \hat{j}+2 \hat{k}$ that has magnitude 9 is

$\int_0^{\pi /2} {\frac{{\sin x\cos x}}{{1 + {{\sin }^4}x}}\,dx = } $

Consider $f(x)=\left\{\begin{matrix}
tan^{-1}(\frac{\alpha x+\beta}{\gamma})\ \ \ x\in(0,\frac{1}{2}) and \\0 \ \ \ \ \ \ \ x=\frac{1}{2}
and \\ ln(\beta x^2 +2) \ \ \ \ \ \ x\in(\frac{1}{2},1)
and\end{matrix}\right.$ . If $f(x)$ continuous and derivable in its domain then the value of $\alpha + \beta + \gamma$ is-

Number of functions $f:[0,1] \rightarrow[0,1]$ satisfying $|f(x)-f(y)|=|x-y|$ for all $x, y$ in $[0,1]$ is