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Let overrightarrow and overrightarrow d be vectors such that |overrightarrow +overrightarrow d|=√29 and overrightarrow ×(2 i+3 j+4 k)=(2 i+3 j+4 k) ×…

JEE Main 2026 — Mathematics Coordinate Geometry

2026mcqhard

Let c→\overrightarrow{\mathrm{c}}c and d→\overrightarrow{\mathrm{d}}d be vectors such that ∣c→+d→∣=29|\overrightarrow{\mathrm{c}}+\overrightarrow{\mathrm{d}}|=\sqrt{29}∣c+d∣=29​ and c→×(2i^+3j^+4k^)=(2i^+3j^+4k^)×d→\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}}c×(2i^+3j^​+4k^)=(2i^+3j^​+4k^)×d. If λ1,λ2(λ1>λ2)\lambda_{1}, \lambda_{2}\left(\lambda_{1}>\lambda_{2}\right)λ1​,λ2​(λ1​>λ2​) are the possible values of (c⃗+d⃗)⋅(−7i^+2j^+3k^)(\vec{c}+\vec{d}) \cdot(-7 \hat{i}+2 \hat{j}+3 \hat{k})(c+d)⋅(−7i^+2j^​+3k^), then the equation K2x2+(K2−5 K+λ1)xy+(3 K+λ22)y2−8x+12y+λ2=0\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}=0K2x2+(K2−5 K+λ1​)xy+(3 K+2λ2​​)y2−8x+12y+λ2​=0 represents a circle, for K equal to :

Official previous-year question

Held on 21 Jan 2026 · Verified 6 Jul 2026.

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