JEE Mains · Maths · STD 12 - 10. vector algebra
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 \( \lambda_1, \lambda_2 \) (\(\lambda_1 \)>\(\lambda_2 \)) are the possible values of \( (\vec{c}+\vec{d}).(-7\hat{i}+2\hat{j}+3\hat{k}) \),
then the equation
\( K^{2}x^{2}+(K^{2}-5K+\lambda_{1})xy+(3K+\frac{\lambda_{2}}{2})y^{2}-8x+12y+\lambda_{2}=0 \)
represents a circle, for k equal to:
- A 4
- B 1
- C -1
- D 2
Answer & Solution
Correct Answer
(B) 1
Step-by-step Solution
Detailed explanation
\( |\vec{c}+\vec{d}|=\sqrt{29} \) \( \vec{c}+\vec{d}=\lambda(2\hat{i}+3\hat{j}+4\hat{k}) \) \(\lambda=\pm1 \) \(\lambda(-14+6+12)=4\lambda \), \( \lambda_1 = 4, \lambda_2 = -4 \) \( k^{2}x^{2}+(k^{2}-5k+4)xy+(3k-2)y^{2}-8x+12y-4 =0\) is circle \(k^2-5 k+4=0 \Rightarrow k=1,4\)…
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