Vector Algebra — Practice

Question 1

Find the projection of $\mathbf b=2\hat i+3\hat j+2\hat k$ on $\mathbf a=\hat i+2\hat j+\hat k$.

Accepted Answer

$\mathbf a\cdot\mathbf b=2+6+2=10$ and $\left|\mathbf aight|=\sqrt6$.

Projection $=\dfrac{10}{\sqrt6}$.

Question 2

Find a unit vector perpendicular to both $\hat i+\hat j$ and $\hat j+\hat k$.

Accepted Answer

$(1,1,0)	imes(0,1,1)=(1\cdot1-0\cdot1,\ 0\cdot0-1\cdot1,\ 1\cdot1-1\cdot0)=(1,-1,1)$.

Magnitude $\sqrt3$, so the unit vector is $\dfrac{1}{\sqrt3}\left(\hat i-\hat j+\hat kight)$, or its negative.

Question 3

Accepted Answer

$\left|\mathbf a	imes\mathbf bight|^{2}+\left(\mathbf a\cdot\mathbf bight)^{2}=\left|\mathbf aight|^{2}\left|\mathbf bight|^{2}$.

$36+\left(\mathbf a\cdot\mathbf bight)^{2}=144$, so $\mathbf a\cdot\mathbf b=\pm6\sqrt3$.

Both signs are possible, corresponding to an acute or an obtuse angle.

Question 4

Find the area of the parallelogram whose diagonals are $3\hat i+\hat j-2\hat k$ and $\hat i-3\hat j+4\hat k$.

Accepted Answer

$(3,1,-2)	imes(1,-3,4)=(4-6,\ -2-12,\ -9-1)=(-2,-14,-10)$.

Magnitude $=\sqrt{4+196+100}=\sqrt{300}=10\sqrt3$.

Area $=5\sqrt3$.

Question 5

Show that $\mathbf a=\hat i-2\hat j+3\hat k$, $\mathbf b=-2\hat i+3\hat j-4\hat k$ and $\mathbf c=\hat i-3\hat j+5\hat k$ are coplanar.

Accepted Answer

$\begin{vmatrix}1&-2&3\-2&3&-4\1&-3&5\end{vmatrix}=1(15-12)+2(-10+4)+3(6-3)$.

$=3-12+9=0$.

The triple product vanishes, so the three are coplanar and linearly dependent.

Question 6

Find the volume of the tetrahedron with vertices $(1,1,1)$, $(2,1,3)$, $(3,2,2)$ and $(3,3,4)$.

Accepted Answer

Edges from the first vertex: $(1,0,2)$, $(2,1,1)$, $(2,2,3)$.

$\begin{vmatrix}1&0&2\2&1&1\2&2&3\end{vmatrix}=1(3-2)-0+2(4-2)=1+4=5$.

Volume $=\dfrac56$.

Question 7

Simplify $\mathbf a	imes\left(\mathbf b	imes\mathbf cight)+\mathbf b	imes\left(\mathbf c	imes\mathbf aight)+\mathbf c	imes\left(\mathbf a	imes\mathbf bight)$.

Accepted Answer

$\left(\mathbf a\cdot\mathbf cight)\mathbf b-\left(\mathbf a\cdot\mathbf bight)\mathbf c+\left(\mathbf b\cdot\mathbf aight)\mathbf c-\left(\mathbf b\cdot\mathbf cight)\mathbf a+\left(\mathbf c\cdot\mathbf bight)\mathbf a-\left(\mathbf c\cdot\mathbf aight)\mathbf b$.

Every term cancels in pairs, since the dot product is symmetric.

The sum is $\mathbf 0$.

Question 8

Solve for $\mathbf r$: $\mathbf r	imes\left(\hat i+\hat jight)=\hat k$ and $\mathbf r\cdot\hat i=2$.

Accepted Answer

Write $\mathbf r=(x,y,z)$. Then $\mathbf r	imes(1,1,0)=(-z,\ z,\ x-y)$.

Setting this to $(0,0,1)$ gives $z=0$ and $x-y=1$.

With $x=2$ we get $y=1$, so $\mathbf r=2\hat i+\hat j$.

Question 9

Show that the points $(2,-1,3)$, $(4,3,1)$ and $(3,1,2)$ are collinear.

Accepted Answer

$\overrightarrow{AB}=(2,4,-2)$ and $\overrightarrow{AC}=(1,2,-1)$.

$\overrightarrow{AB}=2\,\overrightarrow{AC}$, so the cross product vanishes.

The three points are collinear, and $C$ is the midpoint of $AB$.

Question 10

Find a vector along the internal bisector of the angle between $\mathbf a=2\hat i+\hat j-2\hat k$ and $\mathbf b=\hat i+2\hat j+2\hat k$.

Accepted Answer

Both have magnitude $3$.

$\hat{\mathbf a}+\hat{\mathbf b}=\dfrac{(2,1,-2)+(1,2,2)}{3}=\dfrac{(3,3,0)}{3}=(1,1,0)$.

So $\hat i+\hat j$ lies along the bisector. Here the two magnitudes happen to be equal, so the unnormalised sum works too — which is exactly why the general case must be checked.

Question 11

Prove that the altitudes of a triangle are concurrent, using vectors.

Accepted Answer

Let $O$ be where the altitudes from $A$ and $B$ meet, with position vectors $\mathbf a,\mathbf b,\mathbf c$ from $O$.

The first altitude gives $\mathbf a\cdot\left(\mathbf b-\mathbf cight)=0$ and the second $\mathbf b\cdot\left(\mathbf c-\mathbf aight)=0$.

Adding, $\mathbf a\cdot\mathbf b-\mathbf a\cdot\mathbf c+\mathbf b\cdot\mathbf c-\mathbf b\cdot\mathbf a=0$, so $\mathbf c\cdot\left(\mathbf b-\mathbf aight)=0$.

That is the third altitude, so it passes through $O$ too.

Question 12

Find the moment about the point $(1,0,0)$ of a force $\hat i+2\hat j-\hat k$ acting at $(2,1,1)$.

Accepted Answer

$\mathbf r=(2,1,1)-(1,0,0)=(1,1,1)$.

$\mathbf r	imes\mathbf F=(1,1,1)	imes(1,2,-1)=(-1-2,\ 1+1,\ 2-1)=(-3,2,1)$.

Practice — Vector Algebra

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