Let c a n d d be vectors such that | c + d | = 29 and c × ( 2 i + 3 j + 4 k ) = ( 2 i + 3 j + 4 k ) × d . If 1 , 2 ( 1 2 ) are the possible values of ( c + d ) · ( - 7 i + 2 j + 3 k ) , then the equation K 2 x 2 + ( K 2 - 5 K + 1 ) x y + 3

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Solution

Q.
Let $\vec{c} and \vec{d}$ be vectors such that $| \vec{c} + \vec{d} |$ $= \sqrt{29}$ and $\vec{c} \times (2 \hat{i} + 3 \hat{j} + 4 \hat{k}) = (2 \hat{i} + 3 \hat{j} + 4 \hat{k}) \times \vec{d} .$ If $λ_{1}, λ_{2} (λ_{1} > λ_{2})$ are the possible values of $(\vec{c} + \vec{d}) \cdot (- 7 \hat{i} + 2 \hat{j} + 3 \hat{k}),$ then the equation $K^{2} x^{2} + (K^{2} - 5 K + λ_{1}) x y + (3 K + \frac{λ_{2}}{2}) y^{2} - 8 x + 12 y + λ_{2} = 0$ represents a circle, for k equal to: [2026]

1 -1

2 4

3 1

4 2

Ans.
(3)

$| \vec{c} + \vec{d} |$ $= \sqrt{29}$

$\vec{c} + \vec{d} = λ (2 \hat{i} + 3 \hat{j} + 4 \hat{k})$

$λ = \pm 1$

$λ (- 14 + 6 + 12) = 4 λ, λ_{1} = 4, λ_{2} = - 4$

$k^{2} x^{2} + (k^{2} - 5 k + 4) x y + (3 k - 2) y^{2} - 8 x + 12 y - 4 = 0$

$is circle$

$k^{2} - 5 k + 4 = 0 \Rightarrow k = 1, 4$

$k^{2} = 3 k - 2 \Rightarrow k = 1, 2$

$k = 1$

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