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Vector Algebra question

2025 · 28 Jan · Shift 1 · Q50
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Vector Algebra question

2025 · 28 Jan · Shift 1 · Q50

JEE MainMathematicsVector AlgebraNumerical+4 / −1
Let a⃗=i^+j^+k^,b→=2i^+2j^+k^\vec{a}=\hat{\mathrm{i}}+\hat{\mathrm{j}}+\hat{\mathrm{k}}, \overrightarrow{\mathrm{b}}=2 \hat{\mathrm{i}}+2 \hat{\mathrm{j}}+\hat{\mathrm{k}}a=i^+j^​+k^,b=2i^+2j^​+k^ and d→=a⃗×b→\overrightarrow{\mathrm{d}}=\vec{a} \times \overrightarrow{\mathrm{b}}d=a×b. If c→\overrightarrow{\mathrm{c}}c is a vector such that a⃗⋅c→=∣c→∣\vec{a} \cdot \overrightarrow{\mathrm{c}}=|\overrightarrow{\mathrm{c}}|a⋅c=∣c∣, ∣c→−2a⃗∣2=8|\overrightarrow{\mathrm{c}}-2 \vec{a}|^2=8∣c−2a∣2=8 and the angle between d→\overrightarrow{\mathrm{d}}d and c→\overrightarrow{\mathrm{c}}c is π4\frac{\pi}{4}4π​, then ∣10−3 b→⋅c→∣+∣d→×c→∣2|10-3 \overrightarrow{\mathrm{~b}} \cdot \overrightarrow{\mathrm{c}}|+|\overrightarrow{\mathrm{d}} \times \overrightarrow{\mathrm{c}}|^2∣10−3 b⋅c∣+∣d×c∣2 is equal to ‾\underline{\hspace{2cm}}​.
Numerical answer
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