Reply To: Question 11 — JEE Main 24 Jan 2026 S Ii

shahkb4
3 June, 2026 at 8:15 pm #1050

Solution

Any vector $\overrightarrow{\mathrm{v}}$ in plane of $\overrightarrow{\mathrm{a}} \& \overrightarrow{\mathrm{~b}}$

\[ \begin{aligned} & \vec{v}=x \vec{a}+y \vec{b}=x(2 \hat{\imath}-\hat{j}-\hat{k})+y(\hat{\imath}+3 \hat{j}-\hat{k}) \\ & \Rightarrow \vec{v}=(2 x+y) \hat{\imath}+(3 y-x) \hat{j}-(x+y) \hat{k} \end{aligned} \]

Projection length $\vec{v}$ on $\vec{c}=\frac{|\dot{v} \cdot \vec{c}|}{|\vec{c}|}=\frac{1}{\sqrt{14}}$
$\Rightarrow \frac{|2(2 x+y)+1(3 y-x)-7(x+y)|}{\sqrt{4+1+9}}=\frac{1}{\sqrt{14}}$
$\Rightarrow|4 x+2 y+3 y-x-3 x-3 y|=1$
$\Rightarrow|2 \mathrm{y}|=1 \Rightarrow \mathrm{y}= \pm \frac{1}{2}$
$|\overrightarrow{\mathrm{V}}|=\sqrt{(2 \mathrm{x}+\mathrm{y})^{2}+(3 \mathrm{y}-\mathrm{x})^{2}+(\mathrm{x}+\mathrm{y})^{2}}$
$=\sqrt{4 x^{2}+y^{2}+4 x y+9 y^{2}+x^{2}-6 x y+x^{2}+y^{2}+2 x y}$
$=\sqrt{6 x^{2}+11 y^{2}}=\sqrt{6 x^{2}+\frac{11}{4}}=\frac{\sqrt{24 x^{2}+11}}{2}$
(1) $\frac{\sqrt{24 x^{2}+11}}{2}=\frac{\sqrt{21}}{2} \Rightarrow 24 x^{2}+11=21$
$\Rightarrow \mathrm{x}^{2}=\frac{10}{24} \Rightarrow \mathrm{x}= \pm \sqrt{\frac{5}{12}}$
(2) $\frac{\sqrt{24 \mathrm{x}^{2}+11}}{2}=13 \Rightarrow 24 \mathrm{x}^{2}+11=169 \times 4$
$\Rightarrow \mathrm{x} \approx \pm 5.26$
(3) $\frac{\sqrt{24 x^{2}+11}}{2}=\frac{\sqrt{35}}{2} \Rightarrow x=1$
(4) $\frac{\sqrt{24 \mathrm{x}^{2}+11}}{2}=7 \Rightarrow 24 \mathrm{x}^{2}+11=4 \times 49$
$\Rightarrow \mathrm{x} \approx \pm 2.27$ Answer(1,2,3,4)