IMO Practice Test — Vector Algebra
6 Questions • 20 min • Olympiad level
20:00
Question 1 of 6
medium
If $|\vec{a}+\vec{b}|=|\vec{a}-\vec{b}|$, then:
$\vec{a}=\vec{b}$
$\vec{a}\perp\vec{b}$
$|\vec{a}|=|\vec{b}|$
$\vec{a}=2\vec{b}$
Explanation: $|\vec{a}+\vec{b}|^2=|\vec{a}-\vec{b}|^2 \Rightarrow 4\vec{a}\cdot\vec{b}=0 \Rightarrow \vec{a}\perp\vec{b}$.
Question 2 of 6
hard
$\vec{a}\times(\vec{b}\times\vec{c})=$
$(\vec{a}\cdot\vec{c})\vec{b}-(\vec{a}\cdot\vec{b})\vec{c}$
$(\vec{a}\cdot\vec{b})\vec{c}-(\vec{a}\cdot\vec{c})\vec{b}$
$\vec{b}(\vec{a}\cdot\vec{c})+\vec{c}(\vec{a}\cdot\vec{b})$
$(\vec{a}\times\vec{b})\times\vec{c}$
Explanation: Vector triple product: $\vec{a}\times(\vec{b}\times\vec{c})=(\vec{a}\cdot\vec{c})\vec{b}-(\vec{a}\cdot\vec{b})\vec{c}$ (BAC-CAB rule).
Question 3 of 6
hard
If $\vec{a},\vec{b},\vec{c}$ are unit vectors and $\vec{a}+\vec{b}+\vec{c}=\vec{0}$, then $\vec{a}\cdot\vec{b}+\vec{b}\cdot\vec{c}+\vec{c}\cdot\vec{a}=$
$-3/2$
$0$
$3$
$-1$
Explanation: $|\vec{a}+\vec{b}+\vec{c}|^2=0\Rightarrow 3+2(\vec{a}\cdot\vec{b}+\vec{b}\cdot\vec{c}+\vec{c}\cdot\vec{a})=0\Rightarrow$ sum $=-3/2$.
Question 4 of 6
medium
Vectors $\vec{a},\vec{b},\vec{c}$ are coplanar iff:
$\vec{a}\cdot(\vec{b}\times\vec{c})=0$
$\vec{a}\times\vec{b}=\vec{0}$
$\vec{a}\cdot\vec{b}=0$
$|\vec{a}|=|\vec{b}|=|\vec{c}|$
Explanation: Scalar triple product $[\vec{a},\vec{b},\vec{c}]=\vec{a}\cdot(\vec{b}\times\vec{c})=0$ iff the three vectors are coplanar.
Question 5 of 6
hard
If $|\vec{a}|=10$, $|\vec{b}|=2$, $|\vec{a}\times\vec{b}|=16$, then $\vec{a}\cdot\vec{b}=$
$12$
$16$
$4$
$8$
Explanation: $|\vec{a}\times\vec{b}|=|\vec{a}||\vec{b}|\sin\theta=20\sin\theta=16$, $\sin\theta=4/5$, $\cos\theta=3/5$. $\vec{a}\cdot\vec{b}=20\times3/5=12$.
Question 6 of 6
hard
The shortest distance between the lines $\vec{r}=\vec{a}_1+\lambda\vec{b}$ and $\vec{r}=\vec{a}_2+\mu\vec{b}$ (parallel lines) is:
$\frac{|(\vec{a}_2-\vec{a}_1)\times\vec{b}|}{|\vec{b}|}$
$\frac{|\vec{a}_2-\vec{a}_1|}{|\vec{b}|}$
$|(\vec{a}_2-\vec{a}_1)|$
$\frac{|\vec{b}|}{|\vec{a}_2-\vec{a}_1|}$
Explanation: For parallel lines with the same direction $\vec{b}$, the shortest distance (perpendicular distance) is $\frac{|(\vec{a}_2-\vec{a}_1)\times\vec{b}|}{|\vec{b}|}$.