Q27 / 75 · JEE Main 2026 (21 January Shift 1) · Mathematics
Let $\overrightarrow{\mathrm{c}}$ and $\overrightarrow{\mathrm{d}}$ be vectors such that $|\overrightarrow{\mathrm{c}}+\overrightarrow{\mathrm{d}}|=\sqrt{29}$ and $\overrightarrow{\mathrm{c}} \times(2 \hat{i}+3 \hat{j}+4 \hat{k})=(2 \hat{i}+3 \hat{j}+4 \hat{k}) \times \overrightarrow{\mathrm{d}}$. If $\lambda_{1}, \lambda_{2}\left(\lambda_{1}>\lambda_{2}\right)$ are the possible values of $(\vec{c}+\vec{d}) \cdot(-7 \hat{i}+2 \hat{j}+3 \hat{k})$, then the equation $\mathrm{K}^{2} x^{2}+\left(\mathrm{K}^{2}-5 \mathrm{~K}+\lambda_{1}\right) x y+\left(3 \mathrm{~K}+\frac{\lambda_{2}}{2}\right) y^{2}-8 x+12 y+\lambda_{2}=0$ represents a circle, for K equal to :
- Correct answerA1
- B4
- C-1
- D2
Solution
From $\vec{c} \times \vec{a} = \vec{a} \times \vec{d}$ where $\vec{a} = 2\hat{i} + 3\hat{j} + 4\hat{k}$, we get $(\vec{c} + \vec{d}) \times \vec{a} = 0$.
So $\vec{c} + \vec{d} = t\vec{a}$. Given $|\vec{c} + \vec{d}| = \sqrt{29}$: $|t| \cdot \sqrt{29} = \sqrt{29}$, so $t = \pm 1$.
$(\vec{c} + \vec{d}) \cdot (-7\hat{i} + 2\hat{j} + 3\hat{k}) = \pm(-14 + 6 + 12) = \pm 4$.
$\lambda_1 = 4$, $\lambda_2 = -4$.
For circle: Coeff of $x^2$ = Coeff of $y^2$: $K^2 = 3K - 2 \Rightarrow K = 1, 2$.
Coeff of $xy = 0$: $K^2 - 5K + 4 = 0 \Rightarrow K = 1, 4$.
Common value: $K = 1$.
So $\vec{c} + \vec{d} = t\vec{a}$. Given $|\vec{c} + \vec{d}| = \sqrt{29}$: $|t| \cdot \sqrt{29} = \sqrt{29}$, so $t = \pm 1$.
$(\vec{c} + \vec{d}) \cdot (-7\hat{i} + 2\hat{j} + 3\hat{k}) = \pm(-14 + 6 + 12) = \pm 4$.
$\lambda_1 = 4$, $\lambda_2 = -4$.
For circle: Coeff of $x^2$ = Coeff of $y^2$: $K^2 = 3K - 2 \Rightarrow K = 1, 2$.
Coeff of $xy = 0$: $K^2 - 5K + 4 = 0 \Rightarrow K = 1, 4$.
Common value: $K = 1$.
Hence, the correct answer is option: 'A'.