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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 $\vec a = \hat i - \hat j,$ $\vec b = \hat i + \hat j + \hat k$ and $\vec c$ be a vector such that $\vec a \times \vec c + \vec b = 0$ and $\vec a.\vec c = 4$, then ${\left| {\vec c} \right|^2}$ is equal to
  • ✓
    $\frac{{19}}{2}$
  • B
    $9$
  • C
    $8$
  • D
    $\frac{{17}}{2}$
✓

Answer

Correct option: A.
$\frac{{19}}{2}$
a
$a=\hat{i}-\hat{j}, b=\hat{i}+\hat{j}+\hat{k}, c=x \hat{i}+y \hat{j}+z \hat{k}$

$\vec{a} \times \vec{c}+\vec{b}=0$

$ \Rightarrow \left| {\begin{array}{*{20}{c}}
{\hat i}&{\hat j}&{\hat k}\\
1&{ - 1}&0\\
x&y&z
\end{array}} \right|$

$+(\hat{i}+\hat{\bar{j}}+\hat{k})=0$

$\hat{i}(-z)-\hat{j}(z)+\hat{k}(y+x)$

$\Rightarrow 1-z=0 \Rightarrow z=1$

$\text { Also } x+y=-1, \text { and } \vec{a} \cdot \vec{c}=4 \Rightarrow x-y=4$

$\Rightarrow x=\frac{3}{2}, y=\frac{5}{2}$

$\therefore|\overrightarrow{\mathrm{c}}|^{2}=x^{2}+y^{2}+z^{2}$

$=\frac{9}{4}+\frac{25}{4}+1=\frac{38}{4}=\frac{19}{2}$

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