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10. vector algebra — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMaths10. vector algebra1 Mark
MCQ
Let $\vec{a}=4 \hat{i}-\hat{j}+\hat{k}, \vec{b}=11 \hat{i}-\hat{j}+\hat{k}$ and $\vec{c}$ be a vector such that $(\vec{a}+\vec{b}) \times \vec{c}=\vec{c} \times(-2 \vec{a}+3 \vec{b}) \text {. }$

If $(2 \vec{a}+3 \vec{b}) \cdot \vec{c}=1670$, then $|\vec{c}|^2$ is equal to :

  • A
    $1627$
  • ✓
    $1618$
  • C
    $1600$
  • D
    $1609$
✓

Answer

Correct option: B.
$1618$
b
$ (\vec{a}+\vec{b}) \times \vec{c}-\vec{c} \times(-2 \vec{a}+3 \vec{b})=0 $

$ (\vec{a}+\vec{b}) \times \vec{c}+(-2 \vec{a}+3 \vec{b}) \times \vec{c}=0 $

$ \Rightarrow(\vec{a}+\vec{b})-2 \vec{a}+3 \vec{b}) \times \vec{c}=0 $

$ \Rightarrow \vec{c}=\lambda(4 \vec{b}-\vec{a}) $

$ \Rightarrow=\lambda(44 \hat{i}-4 \hat{j}+4 \hat{k}-4 \hat{i}+\hat{j}-\hat{k}) $

$ =\lambda(40 \hat{i}-3 \hat{j}+3 \hat{k})$

Now

$ (8 \hat{i}-2 \hat{j}+2 \hat{k}+33 \hat{i}-3 \hat{j}+3 \hat{k}) \cdot \lambda(40 \hat{i}-3 \hat{j}+3 \hat{k})=1670 $

$ \Rightarrow(41 \hat{i}-5 \hat{j}+5 \hat{k}) \cdot(40 \hat{i}-3 \hat{j}+3 \hat{k}) \times \lambda=1670) $

$ \Rightarrow(1640+15+15) \lambda=1670 \Rightarrow \lambda=1 $

$ \text { so } \overrightarrow{\mathrm{c}}=40 \hat{i}-3 \hat{j}-3 \hat{k} $

$ \Rightarrow|\overrightarrow{\mathrm{c}}|^2=1600+9+9=1618$

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