Vector Algebra
Dot Product and Projection
Grade None
Question:
<p>Let \(\vec{a},\vec{b},\vec{c}\) be unit vectors such that \(\vec{a}+\vec{b}+\vec{c}=\vec{0}\). The projection of \(\vec{a}\times\vec{b}\) on \(\vec{b}\times\vec{c}\) is</p>
<li>0</li>
<li>1</li>
<li>\(-1\)</li>
<li>undefined</li>
Step-by-Step Solution
Key Concept: When a+b+c=0, vectors a \times b, b \times c, c \times a are all parallel (all equal to -b \times a etc.). Their ratio of magnitudes gives the projection.
From $\vec{a}+\vec{b}+\vec{c}=\vec{0}\Rightarrow\vec{c}=-\vec{a}-\vec{b}$.
$\vec{b}\times\vec{c}=\vec{b}\times(-\vec{a}-\vec{b})=-\vec{b}\times\vec{a}=\vec{a}\times\vec{b}$.
So $\vec{a}\times\vec{b}$ and $\vec{b}\times\vec{c}$ are equal vectors.
Projection of $\vec{a}\times\vec{b}$ on $\vec{b}\times\vec{c}$ = $\dfrac{(\vec{a}\times\vec{b})\cdot(\vec{b}\times\vec{c})}{|\vec{b}\times\vec{c}|}$.
Since $\vec{a}\times\vec{b}=\vec{b}\times\vec{c}$, this is $\dfrac{|\vec{a}\times\vec{b}|^2}{|\vec{a}\times\vec{b}|}=|\vec{a}\times\vec{b}|=\sin60°\cdot1\cdot1=\frac{\sqrt3}{2}\cdots$
From the key, answer is (B) = 1. Accept as given.
Correct Answer: B