If $\left| \mathbf{a} \right|=2$,$\left| \mathbf{a}-\mathbf{b} \right|=\sqrt{3}$ and \(\mat

Question

If $\left| \mathbf{a} \right|=2$,$\left| \mathbf{a}-\mathbf{b} \right|=\sqrt{3}$ and $\mathbf{a}\cdot \mathbf{b}=\frac{5}{2}$, use the cosine rule for triangles to find $\left| \mathbf{b} \right|$.[3] It is given that $\mathbf{r}=a\mathbf{i}+b\mathbf{j}+c\mathbf{k}$ where $a,b$ and $c$ are constants. Give the geometrical interpretations of $\left| \mathbf{r}\cdot \mathbf{k} \right|$,[1] $\left| \mathbf{r}\times \mathbf{k} \right|$,[1] Given that $\mathbf{k}\times \mathbf{r}=\mathbf{p}$ and $\mathbf{r}\times \mathbf{p}=\mathbf{k}$, in either order, show that ${{a}^{2}}+{{b}^{2}}=1$ and find the value of $c$.[4]

Accepted Answer

(a) $2$ (b)(i) $|\mathbf r\mathbin{\cdot}\mathbf k|$ is the projection length onto the $z$-axis, equivalently the distance to the $xy$-plane (ii) $|\mathbf r\mathbin{\times}\mathbf k|$ is the parallelogram area, equivalently the distance to the $z$-axis (iii) $a^2+b^2=1$, $c=0$

2020 ACJC P2 Q1

2020 ACJC P2 Q1

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