Porisma

Worked solution

Use cross-product antisymmetry before taking a magnitude

Given ∣a×b∣=4|\mathbf a\times\mathbf b|=4, find the value of ∣(a+2b)×(3a−b)∣|(\mathbf a+2\mathbf b)\times(3\mathbf a-\mathbf b)|.

CJC · Discussion · Question 9

Answer

2828.

Expand in the printed order

(a+2b)×(3a−b)=3a×a+6b×a−a×b−2b×b=−7a×b.\begin{aligned}(\mathbf a+2\mathbf b)\times(3\mathbf a-\mathbf b)&=3\mathbf a\times\mathbf a+6\mathbf b\times\mathbf a-\mathbf a\times\mathbf b-2\mathbf b\times\mathbf b\\&=-7\mathbf a\times\mathbf b.\end{aligned}

Here a×a=b×b=0\mathbf a\times\mathbf a=\mathbf b\times\mathbf b=\mathbf 0 and b×a=−a×b\mathbf b\times\mathbf a=-\mathbf a\times\mathbf b.

How did this part go?

Take the nonnegative magnitude

∣−7a×b∣=7∣a×b∣=7(4)=28|-7\mathbf a\times\mathbf b|=7|\mathbf a\times\mathbf b|=7(4)=28. The intermediate vector is a negative multiple; its magnitude is positive.

How did this part go?

Common mistakes

The source asks for a magnitude. The answer is 28, not a vector and not −28.

Solution notes

All printed cross-product terms and magnitude bars restored from the page image.

Read the hint