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CBSE Class 12 Math 2022 (Term II) Delhi Set 1 Solved Paper

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Question : 4 of 14
Marks: +1, -0
If a⃗=i^+j^+k^,a⃗⋅b⃗=1\vec{a}=\hat{i}+\hat{j}+\hat{k}, \vec{a}\cdot \vec{b}=1 and a⃗×b⃗=j^−k^\vec{a}\times \vec{b}=\hat{j}-\hat{k}, then find ∣b⃗∣\left|\vec{b}\right|
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