Vector Algebra
Cross product and dot product
Grade 12
Question:
<p>Let \(\vec{a}=\hat{i}+\hat{j}+\hat{k}\), \(\vec{c}=\hat{j}-\hat{k}\) and a vector \(\vec{b}\) be such that \(\vec{a}\times\vec{b}=\vec{c}\) and \(\vec{a}\cdot\vec{b}=3\). Then \(|\vec{b}|\) equals</p>
<p>\(\dfrac{11}{3}\)</p>
<p>\(\dfrac{11}{\sqrt{3}}\)</p>
<p>\(\sqrt{\dfrac{11}{3}}\)</p>
<p>\(\dfrac{\sqrt{11}}{3}\)</p>
Step-by-Step Solution
Key Concept: Use the two constraints simultaneously: the cross product equation determines the perpendicular component of b, while the dot product gives the parallel component. Combine using |b|² = (b∥)² + (b⊥)².
Step 1: From a · b = 3, the component of b parallel to a is: b∥ = (a·b/|a|^2) a = (3/3) a = a Step 2: From a × b = c , take the magnitude: | a × b | = | c | | a || b |sin(θ) = √(1+1) = √2, so √3·| b |sin(θ) = √2 Step 3: The perpendicular component satisfies | b ⊥|^2 = | a × b |^2/| a |^2 = 2/3 Step 4: Since b = b ∥ + b ⊥: | b |^2 = | b ∥|^2 + | b ⊥|^2 = | a |^2 + 2/3 = 3 + 2/3 = 11/3 ∴ | b | = √(11/3) = √33/3 Answer: C
Correct Answer: C