Download App

STD 12 - 10. vector algebra — Mathematics STD 12 Science — Question

CBSE BoardEnglish MediumSTD 12 ScienceMathematicsSTD 12 - 10. vector algebra1 Mark
MCQ
Let $\overrightarrow{\mathrm{a}}$ and $\overrightarrow{\mathrm{b}}$ be two vectors such that $|2 \vec{a}+3 \vec{b}|=|3 \vec{a}+\vec{b}|$ and the angle between $\vec{a}$ and $\vec{b}$ is $60^{\circ}$. If $\frac{1}{8} \vec{a}$ is a unit vector, then $|\vec{b}|$ is equal to :
  • A
    $4$
  • B
    $6$
  • ✓
    $5$
  • D
    $8$
✓

Answer

Correct option: C.
$5$
c
$|3 \vec{a}+\vec{b}|^{2}=|2 \vec{a}+3 \vec{b}|^{2}$

$(3 \vec{a}+\vec{b}) \cdot(3 \vec{a}+\vec{b})=(2 \vec{a}+3 \vec{b}) \cdot(2 \vec{a}+3 \vec{b})$

$9 \vec{a} \cdot \vec{a}+6 \vec{a} \cdot \vec{b}+\vec{b} \cdot \vec{b}=4 \vec{a} \cdot \vec{a}+12 \vec{a} \cdot \vec{b}+9 \cdot \vec{b} \cdot \vec{b}$

$5|\vec{a}|^{2}-6 \vec{a} \cdot \vec{b}=8|\vec{b}|^{2}$

$5(8)^{2}-6.8 .|\vec{b}| \cos 60^{\circ}=8|\vec{b}|^{2}$ $[\frac{1}{8}|\vec{a}|=1\Rightarrow|\vec{a}|=8]$

$40-3|\overrightarrow{\mathrm{b}}|=|\overrightarrow{\mathrm{b}}|^{2}$

$\Rightarrow|\overrightarrow{\mathrm{b}}|^{2}+3|\overrightarrow{\mathrm{b}}|-40=0$

$|\overrightarrow{\mathrm{b}}|=-8, \quad|\overrightarrow{\mathrm{b}}|=5$

$\quad$(rejected)

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 STD 12 - 10. vector algebra→STD 12 - 10. vector algebra — all question types→Mathematics STD 12 Science question papers→CBSE Board STD 12 Science question papers→CBSE Board question paper generator→

Similar questions

If $f(x)=|\cos x|$, then $f\left(\frac{3 \pi}{4}\right)$ is
→
If $\text{g(f(x))}=|\sin\text{x}|$ and $\text{f(g(x))}=(\sin\sqrt{\text{x}})^2,$ then
→
The circumference of a circle is measured as 28cm with an error of 0.01cm. The percentage error in the area is:
→
If the sum of two unit vectors is a unit vector, then the magnitude of their difference is
→
Let $f(x) = \left\{ \begin{array}{l}\frac{{{x^3} + {x^2} - 16x + 20}}{{{{(x - 2)}^2}}},{\rm{if }}\;x \ne 2\\\;\;\;\;\;\,\;\;\;\;\;\;\;k\;\;\;\;\;\;\;\;,\;{\rm{if }}\;x = 2\end{array} \right.$ If $f(x)$ be continuous for all $x$, then $ k =$
→
If $f(x)$ is a quadratic expression such that $f(1) + f (2)\, = 0$ , and $-1$ is a root of $f(x)\, = 0$, then the other root of $f(x)\, = 0$ is
→
A normal to the plane x = 2 is:
→
The solution of the differential equation $\frac{{dy}}{{dx}} + \frac{y}{x} = {x^2}$is
→
If $f(x) = \cos x\cos 2x\cos 4x\cos 8x\cos 16x$, then $f'\left( {{\pi \over 4}} \right)$ is
→
The edge of a cube is of length $‘a’$ then the shortest distance between the diagonal of a cube and an edge skew to it is
→