Question 4
The vectors $2\hat{\boldsymbol{\imath}}-3\hat{\boldsymbol{\jmath}}+\hat{\boldsymbol{k}}$ and $m\hat{\boldsymbol{\imath}}+2\hat{\boldsymbol{\jmath}}+8\hat{\boldsymbol{k}}$ are perpendicular.
Determine the value of $m$.
Not quite. You made a sign error in the $\hat{\boldsymbol{\jmath}}$ product, taking $(-3)\times2$ as $+6$. Then $2m+6+8=0$ gives $2m=-14$, so $m=-7$.
That’s it. Step through the hints below to see why.
Not quite. Your equation $2m+2=0$ is right, but you moved the 2 across without changing its sign. That gives $2m=2$, so $m=1$.
Not quite. You left out the $\hat{\boldsymbol{k}}$ product. Then $2m-6=0$ gives $m=3$.
Set the scalar product to zero.
$$(2\hat{\boldsymbol{\imath}}-3\hat{\boldsymbol{\jmath}}+\hat{\boldsymbol{k}})\cdot(m\hat{\boldsymbol{\imath}}+2\hat{\boldsymbol{\jmath}}+8\hat{\boldsymbol{k}})=0$$
Two non-zero vectors are perpendicular exactly when their scalar (dot) product is 0, because $\boldsymbol{a}\cdot\boldsymbol{b}=|\boldsymbol{a}||\boldsymbol{b}|\cos(90^{\circ})=0$.
Multiply matching components: the $\hat{\boldsymbol{\imath}}$ components, the $\hat{\boldsymbol{\jmath}}$ components and the $\hat{\boldsymbol{k}}$ components.
$$\begin{aligned}2\times m&=2m\\(-3)\times2&=-6\\1\times8&=8\end{aligned}$$
The scalar product is the sum of the products of corresponding components. The first vector’s $\hat{\boldsymbol{k}}$ coefficient is $1$, even though it is not printed.
Add the products, set the total equal to zero and solve.
$$\begin{aligned}2m-6+8&=0\\2m+2&=0\\2m&=-2\\m&=-1\end{aligned}$$
Subtract 2 from both sides, then divide by 2.
Check: $2(-1)-6+8=0$.
The scalar product is $2m-6+8=2m+2$. Setting it to zero gives $m=-1$.
Only mark this done when you could do it without help. Reading the solution does not count.
A Tangent Tuition mock exam question. It is not a past QCAA question, and Tangent Tuition is not affiliated with the QCAA.
Spot a mistake on this page? Report it