Let , . If and , then what is the value of ?

. First, evaluate . Since , , and , this simplifies to . Using Lagrange's identity, . So, . Thus, .

Let \vec{p}=\vec{a}-\vec{b}, \vec{q}=\vec{a}+\vec{b}. If |\vec{a}|=|\vec{b}|=2 and \vec{a}\cdot\vec{b}=2, then what is the value of |\vec{p}\times\vec{q}|?

A. \sqrt{3}
B. \sqrt{6}
C. 2\sqrt{3}
D. 4\sqrt{3} ✓

Correct Answer: D. 4\sqrt{3}

Explanation

First, evaluate \vec{p} \times \vec{q} = (\vec{a}-\vec{b}) \times (\vec{a}+\vec{b}) = \vec{a}\times\vec{a} + \vec{a}\times\vec{b} - \vec{b}\times\vec{a} - \vec{b}\times\vec{b}. Since \vec{a}\times\vec{a} = 0, \vec{b}\times\vec{b} = 0, and \vec{b}\times\vec{a} = -\vec{a}\times\vec{b}, this simplifies to 2(\vec{a}\times\vec{b}). Using Lagrange's identity, |\vec{a}\times\vec{b}|^2 = |\vec{a}|^2|\vec{b}|^2 - (\vec{a}\cdot\vec{b})^2 = (4)(4) - 2^2 = 16 - 4 = 12. So, |\vec{a}\times\vec{b}| = \sqrt{12} = 2\sqrt{3}. Thus, |\vec{p}\times\vec{q}| = 2(2\sqrt{3}) = 4\sqrt{3}.

Let \vec{p}=\vec{a}-\vec{b}, \vec{q}=\vec{a}+\vec{b}. If |\vec{a}|=|\vec{b}|=2 and \vec{a}\cdot\vec{b}=2, then what is the value of |\vec{p}\times\vec{q}|?

Explanation

Related questions on Vector Algebra