12 exam-style questions on the vector product, scalar triple product, areas, volumes and 3D geometry. Tap Show answer for full working.
Given \(\mathbf{a}=\mathbf{i}+2\mathbf{j}-\mathbf{k}\) and \(\mathbf{b}=3\mathbf{i}-\mathbf{j}+2\mathbf{k}\), find \(\mathbf{a}\times\mathbf{b}\).
\(\mathbf{a}\times\mathbf{b}=\begin{vmatrix}\mathbf{i}&\mathbf{j}&\mathbf{k}\\1&2&-1\\3&-1&2\end{vmatrix}=\mathbf{i}(4-1)-\mathbf{j}(2+3)+\mathbf{k}(-1-6)=3\mathbf{i}-5\mathbf{j}-7\mathbf{k}\).
Find a unit vector perpendicular to both \(\mathbf{u}=2\mathbf{i}-\mathbf{j}+\mathbf{k}\) and \(\mathbf{v}=\mathbf{i}+3\mathbf{j}-2\mathbf{k}\).
\(\mathbf{u}\times\mathbf{v}=\begin{vmatrix}\mathbf{i}&\mathbf{j}&\mathbf{k}\\2&-1&1\\1&3&-2\end{vmatrix}=\mathbf{i}(2-3)-\mathbf{j}(-4-1)+\mathbf{k}(6+1)=-\mathbf{i}+5\mathbf{j}+7\mathbf{k}\).
Magnitude \(\sqrt{1+25+49}=\sqrt{75}=5\sqrt{3}\). Unit vector \(\pm\dfrac{1}{5\sqrt{3}}(-\mathbf{i}+5\mathbf{j}+7\mathbf{k})\).
The points \(A\), \(B\) and \(C\) have position vectors \(\mathbf{i}\), \(2\mathbf{i}+\mathbf{j}\) and \(\mathbf{i}+3\mathbf{j}+2\mathbf{k}\) respectively. Find the area of triangle \(ABC\).
\(\overrightarrow{AB}=\mathbf{i}+\mathbf{j}\), \(\overrightarrow{AC}=3\mathbf{j}+2\mathbf{k}\).
\(\overrightarrow{AB}\times\overrightarrow{AC}=\begin{vmatrix}\mathbf{i}&\mathbf{j}&\mathbf{k}\\1&1&0\\0&3&2\end{vmatrix}=2\mathbf{i}-2\mathbf{j}+3\mathbf{k}\). Magnitude \(\sqrt{4+4+9}=\sqrt{17}\).
Area \(=\dfrac{1}{2}\sqrt{17}\).
Evaluate the scalar triple product \([\mathbf{a},\mathbf{b},\mathbf{c}]=\mathbf{a}\cdot(\mathbf{b}\times\mathbf{c})\) where \(\mathbf{a}=\mathbf{i}+\mathbf{j}\), \(\mathbf{b}=\mathbf{j}+\mathbf{k}\), \(\mathbf{c}=\mathbf{k}+\mathbf{i}\). Interpret the result geometrically.
\(\mathbf{b}\times\mathbf{c}=\begin{vmatrix}\mathbf{i}&\mathbf{j}&\mathbf{k}\\0&1&1\\1&0&1\end{vmatrix}=\mathbf{i}(1)-\mathbf{j}(-1)+\mathbf{k}(-1)=\mathbf{i}+\mathbf{j}-\mathbf{k}\).
\(\mathbf{a}\cdot(\mathbf{b}\times\mathbf{c})=1+1+0=2\). The volume of the parallelepiped formed by the three vectors is \(2\).
Show that \(\mathbf{a}\times(\mathbf{b}\times\mathbf{c})=(\mathbf{a}\cdot\mathbf{c})\mathbf{b}-(\mathbf{a}\cdot\mathbf{b})\mathbf{c}\). Verify for \(\mathbf{a}=\mathbf{i}\), \(\mathbf{b}=\mathbf{j}\), \(\mathbf{c}=\mathbf{k}\).
This is the standard vector triple product identity (proof via components or geometric argument).
Verification: left side \(\mathbf{i}\times(\mathbf{j}\times\mathbf{k})=\mathbf{i}\times\mathbf{i}=\mathbf{0}\). Right side \((\mathbf{i}\cdot\mathbf{k})\mathbf{j}-(\mathbf{i}\cdot\mathbf{j})\mathbf{k}=\mathbf{0}\). Holds.
The lines \(L_1:\mathbf{r}=\mathbf{i}+2\mathbf{j}+\lambda(\mathbf{i}-\mathbf{j}+2\mathbf{k})\) and \(L_2:\mathbf{r}=3\mathbf{i}+\mathbf{k}+\mu(2\mathbf{i}+\mathbf{j}-\mathbf{k})\) are skew. Find the shortest distance between them.
\(\mathbf{d}_1=(1,-1,2)\), \(\mathbf{d}_2=(2,1,-1)\). \(\mathbf{d}_1\times\mathbf{d}_2=\begin{vmatrix}\mathbf{i}&\mathbf{j}&\mathbf{k}\\1&-1&2\\2&1&-1\end{vmatrix}=\mathbf{i}(1-2)-\mathbf{j}(-1-4)+\mathbf{k}(1+2)=-\mathbf{i}+5\mathbf{j}+3\mathbf{k}\).
Connecting vector \(\mathbf{a}_2-\mathbf{a}_1=(2,-2,-1)\). Distance \(=\dfrac{|(\mathbf{a}_2-\mathbf{a}_1)\cdot(\mathbf{d}_1\times\mathbf{d}_2)|}{|\mathbf{d}_1\times\mathbf{d}_2|}=\dfrac{|-2-10-3|}{\sqrt{1+25+9}}=\dfrac{15}{\sqrt{35}}=\dfrac{3\sqrt{35}}{7}\).
Find the volume of the tetrahedron with vertices at \(O\), \(A(1,2,0)\), \(B(0,1,3)\) and \(C(2,0,1)\).
Volume \(=\dfrac{1}{6}|[\mathbf{a},\mathbf{b},\mathbf{c}]|=\dfrac{1}{6}\left|\begin{vmatrix}1&2&0\\0&1&3\\2&0&1\end{vmatrix}\right|=\dfrac{1}{6}|1(1-0)-2(0-6)+0|=\dfrac{1}{6}|1+12|=\dfrac{13}{6}\).
Given that \(\mathbf{a}\times\mathbf{b}=\mathbf{c}\times\mathbf{d}\) and \(\mathbf{a}\cdot\mathbf{c}=\mathbf{b}\cdot\mathbf{d}\), show that \((\mathbf{a}-\mathbf{d})\times(\mathbf{b}-\mathbf{c})=\mathbf{0}\) does not necessarily hold, but find a condition under which \(\mathbf{a}-\mathbf{d}\) is parallel to \(\mathbf{b}-\mathbf{c}\).
From \(\mathbf{a}\times\mathbf{b}=\mathbf{c}\times\mathbf{d}\) we have \(\mathbf{a}\times\mathbf{b}-\mathbf{c}\times\mathbf{d}=\mathbf{0}\). Expanding \((\mathbf{a}-\mathbf{d})\times(\mathbf{b}-\mathbf{c})=\mathbf{a}\times\mathbf{b}-\mathbf{a}\times\mathbf{c}-\mathbf{d}\times\mathbf{b}+\mathbf{d}\times\mathbf{c}\) requires the given scalar condition and further assumptions to conclude parallelism. In general the cross product need not vanish; a sufficient condition is that the four vectors are coplanar in a suitable sense.
The plane \(\Pi\) passes through the points \(A(1,0,1)\), \(B(2,1,0)\) and \(C(0,1,2)\). Find a Cartesian equation of \(\Pi\). Find the perpendicular distance from the point \(D(3,3,3)\) to \(\Pi\).
\(\overrightarrow{AB}=(1,1,-1)\), \(\overrightarrow{AC}=(-1,1,1)\). Normal \(\mathbf{n}=\overrightarrow{AB}\times\overrightarrow{AC}=\begin{vmatrix}\mathbf{i}&\mathbf{j}&\mathbf{k}\\1&1&-1\\-1&1&1\end{vmatrix}=2\mathbf{i}+0\mathbf{j}+2\mathbf{k}=(2,0,2)\).
Plane: \(2(x-1)+2(z-1)=0\Rightarrow x+z=2\).
Distance from \(D\): \(\dfrac{|3+3-2|}{\sqrt{1+0+1}}=\dfrac{4}{\sqrt{2}}=2\sqrt{2}\).
Prove that \(|\mathbf{a}\times\mathbf{b}|^2=|\mathbf{a}|^2|\mathbf{b}|^2-(\mathbf{a}\cdot\mathbf{b})^2\). Deduce that \(|\mathbf{a}\times\mathbf{b}|\le|\mathbf{a}||\mathbf{b}|\).
Left side: \(|\mathbf{a}|^2|\mathbf{b}|^2\sin^2\theta\). Right side: \(|\mathbf{a}|^2|\mathbf{b}|^2-|\mathbf{a}|^2|\mathbf{b}|^2\cos^2\theta=|\mathbf{a}|^2|\mathbf{b}|^2\sin^2\theta\). Equal.
Since \(\sin^2\theta\le1\), \(|\mathbf{a}\times\mathbf{b}|\le|\mathbf{a}||\mathbf{b}|\), with equality when \(\theta=\pm\pi/2\).
Three vectors \(\mathbf{a}\), \(\mathbf{b}\) and \(\mathbf{c}\) satisfy \(\mathbf{a}+\mathbf{b}+\mathbf{c}=\mathbf{0}\). Show that \(\mathbf{a}\times\mathbf{b}=\mathbf{b}\times\mathbf{c}=\mathbf{c}\times\mathbf{a}\). Interpret this result for a triangle.
From \(\mathbf{c}=-(\mathbf{a}+\mathbf{b})\), \(\mathbf{a}\times\mathbf{b}=\mathbf{a}\times\mathbf{b}\). \(\mathbf{b}\times\mathbf{c}=\mathbf{b}\times(-(\mathbf{a}+\mathbf{b}))=-\mathbf{b}\times\mathbf{a}-\mathbf{b}\times\mathbf{b}=\mathbf{a}\times\mathbf{b}\). Similarly \(\mathbf{c}\times\mathbf{a}=\mathbf{a}\times\mathbf{b}\).
For a triangle with sides represented by \(\mathbf{a}\), \(\mathbf{b}\), \(\mathbf{c}\) (closed), the cross products are equal and each has magnitude twice the area of the triangle (and the same direction normal to the plane).
Find the equation of the plane which contains the line of intersection of the planes \(x+y+z=1\) and \(2x-y+3z=4\), and which is parallel to the vector \(\mathbf{i}-2\mathbf{j}+\mathbf{k}\).
The required plane has equation \((x+y+z-1)+\lambda(2x-y+3z-4)=0\), i.e. \((1+2\lambda)x+(1-\lambda)y+(1+3\lambda)z-(1+4\lambda)=0\).
It is parallel to \(\mathbf{d}=(1,-2,1)\), so the normal is perpendicular to \(\mathbf{d}\): \((1+2\lambda)(1)+(1-\lambda)(-2)+(1+3\lambda)(1)=0\).
\(1+2\lambda-2+2\lambda+1+3\lambda=0\Rightarrow 5\lambda=0\Rightarrow\lambda=0\). Plane: \(x+y+z=1\). (In this special case the first plane is already parallel to the given vector; check: \(1-2+1=0\), yes.)