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10. vector algebra — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMaths10. 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)

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