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

CBSE BoardEnglish MediumSTD 12 ScienceMathematicsSTD 12 - 10. vector algebra1 Mark
MCQ
Let $\vec{a}=\hat{i}+2 \hat{j}+\lambda \hat{k}, \vec{b}=3 \hat{i}-5 \hat{j}-\lambda \hat{k}, \vec{a} \cdot \vec{c}=7$, $2 \vec{b} \cdot \vec{c}+43=0, \vec{a} \times \vec{c}=\vec{b} \times \vec{c}$. Then $|\vec{a} \cdot \vec{b}|$ is equal to
  • A
    $4$
  • B
    $2$
  • C
    $6$
  • ✓
    $8$
✓

Answer

Correct option: D.
$8$
d
$\overrightarrow{ a }=\hat{ i }+2 \hat{ j }+\lambda \hat{ k }, \quad \overrightarrow{ b }=3 \hat{ i }-5 \hat{ j }-\lambda \hat{ k }, \overrightarrow{ a } \cdot \overrightarrow{ c }=7$

$\overrightarrow{ a } \times \overrightarrow{ c }-\overrightarrow{ b } \times \overrightarrow{ c }=\overrightarrow{0}$

$(\overrightarrow{ a }-\overrightarrow{ b }) \times \overrightarrow{ c }=0 \Rightarrow(\overrightarrow{ a }-\overrightarrow{ b }) \text { is paralleled to } \overrightarrow{ c }$

$\overrightarrow{ a }-\overrightarrow{ b }=\mu \overrightarrow{ c } \text {, where } \mu \text { is a scalar }$

$-2 \hat{ i }+7 \hat{ j }+2 \lambda \hat{ k }=\mu \cdot \overrightarrow{ c }$

Now $\vec{a} \cdot \overrightarrow{ c }=7$ gives $2 \lambda^2+12=7 \mu$

And $\vec{b} \cdot \vec{c}=-\frac{43}{2}$ gives $4 \lambda^2+82=43 \mu$

$\mu=2$ and $\lambda^2=1$

$|\vec{a} \cdot \vec{b}|=8$

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