<p>Vectors $\mathbf{a}$, $\mathbf{b}$ and $\mathbf{c}$ are such that&nbsp;$\mathbf{a}\ne \mathbf{0}$ and $\mathbf{a}\times 3\mathbf{b}=2\mathbf{a}\times \mathbf{c}$.</p>
<ol class="i-list"><li><div class="list-item-paragraph">Show that&nbsp;$3\mathbf{b}-2\mathbf{c}=\lambda \mathbf{a}$, where&nbsp;$\lambda $ is a constant.<span class="inline-marks">[2]</span></div></li><li><div class="list-item-paragraph">It is now given that&nbsp;$\mathbf{a}$ and $\mathbf{c}$ are unit vectors, that the modulus of $\mathbf{b}$ is $4$ and that the angle between $\mathbf{b}$ and $\mathbf{c}$ is $60{}^\circ $. Using a suitable scalar product, find exactly the two possible values of $\lambda $.<span class="inline-marks">[5]</span></div></li></ol>

Vectors $\mathbf{a}$, $\mathbf{b}$ and $\mathbf{c}$ are such that $\mathbf{a}\ne \mathbf{0}$ and $\mathbf{a}\times 3\mathbf{b}=2\mathbf{a}\times \mathbf{c}$.

Show that $3\mathbf{b}-2\mathbf{c}=\lambda \mathbf{a}$, where $\lambda $ is a constant.

It is now given that $\mathbf{a}$ and $\mathbf{c}$ are unit vectors, that the modulus of $\mathbf{b}$ is $4$ and that the angle between $\mathbf{b}$ and $\mathbf{c}$ is $60{}^\circ $. Using a suitable scalar product, find exactly the two possible values of $\lambda $.

Vectors - General

Vectors

H2 Math

International Baccalaureate Higher Level Mathematics

IB HL Math

N2018 P1 Q6

$\mathbf{a} \times 3\mathbf{b} = 2\mathbf{a} \times \mathbf{c}$

Take note: $\mathbf{m} \times \mathbf{n} = \mathbf{0} \Rightarrow \mathbf{m} \parallel \mathbf{n}$

$|\mathbf{m}||\mathbf{n}|\sin\theta \cdot \hat{\mathbf{n}} = \mathbf{0}$ when $\theta = 0^\circ$

$\mathbf{a} \times 3\mathbf{b} - 2\mathbf{a} \times \mathbf{c} = \mathbf{0}$

$\mathbf{a} \times [3\mathbf{b} - 2\mathbf{c}] = \mathbf{0}$

$\Rightarrow \mathbf{a}$ is parallel to $3\mathbf{b} - 2\mathbf{c}$

$\Rightarrow 3\mathbf{b} - 2\mathbf{c} = \lambda\mathbf{a}$ where $\lambda$ is a scalar constant.

$|\mathbf{a}| = |\mathbf{c}| = 1$, $|\mathbf{b}| = 4$, $\angle BOC = 60^\circ$

$\mathbf{a} \cdot \mathbf{a} = |\mathbf{a}|^2$, $\mathbf{c} \cdot \mathbf{c} = |\mathbf{c}|^2$

$3\mathbf{b} - 2\mathbf{c} = \lambda\mathbf{a}$

$(3\mathbf{b} - 2\mathbf{c}) \cdot (3\mathbf{b} - 2\mathbf{c}) = (\lambda\mathbf{a}) \cdot (\lambda\mathbf{a})$

$9\mathbf{b} \cdot \mathbf{b} - 12\mathbf{b} \cdot \mathbf{c} + 4\mathbf{c} \cdot \mathbf{c} = \lambda^2\mathbf{a} \cdot \mathbf{a}$

$9|\mathbf{b}|^2 - 12|\mathbf{b}||\mathbf{c}|\cos 60^\circ + 4|\mathbf{c}|^2 = \lambda^2|\mathbf{a}|^2$

$9(4)^2 - 12(4)(1)\left(\dfrac{1}{2}\right) + 4(1)^2 = \lambda^2(1)^2$

$\lambda = \pm\sqrt{124} = \pm 2\sqrt{31}$

N2018 P1 Q6

N2018 P1 Q6

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