A lvl H2 Maths: Vectors

TPJC 2000 P2 Q14

Referred to the origin O, the points A, B and C have position vectors given respectively by
$\vec{OA}=\bigl(\begin{smallmatrix} 0\\ 0\\ 1 \end{smallmatrix}\bigr), \vec{OB}=\bigl(\begin{smallmatrix} 1\\ -2\\ 3 \end{smallmatrix}\bigr),\vec{OC}=\bigl(\begin{smallmatrix} -2\\ 1\\ 4 \end{smallmatrix}\bigr).$

Find
(i) the cartesian equations of the line l passing through A and B;
(ii) the length of projection of BC on l;
(iii) the image of C in the line l;
(iv) the points where l meets the plane z = 0;
(v) the 2 points P on l such that cos ∡POB = $\frac{2}{\sqrt{7}}$

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Answer:

(i)
$\vec{AB}=\vec{OB}-\vec{OA}$
$=\bigl(\begin{smallmatrix} 1\\ -2\\ 3 \end{smallmatrix}\bigr) -\bigl(\begin{smallmatrix} 0\\ 0\\ 1 \end{smallmatrix}\bigr) =\bigl(\begin{smallmatrix} 1\\ -3\\ 2 \end{smallmatrix}\bigr)$

Hence, line l :
$\bigl(\begin{smallmatrix} 0\\ 1\\ 1 \end{smallmatrix}\bigr)+\lambda \bigl(\begin{smallmatrix} 1\\ -3\\ 2 \end{smallmatrix}\bigr)$

(ii)
$\vec{BC}=\vec{OC}-\vec{OB}$
$=\bigl(\begin{smallmatrix} -2\\ 1\\ 4 \end{smallmatrix}\bigr)- \bigl(\begin{smallmatrix} 1\\ -2\\ 3 \end{smallmatrix}\bigr)= \bigl(\begin{smallmatrix} -3\\ 3\\ 1 \end{smallmatrix}\bigr)$

Thus, length of projection
$=\frac{\left |\bigl(\begin{smallmatrix} -3\\ 3\\ 1 \end{smallmatrix}\bigr)\cdot \bigl(\begin{smallmatrix} 1\\ -3\\ 2 \end{smallmatrix}\bigr) \right |}{\left |\bigl(\begin{smallmatrix} 1\\ -3\\ 2 \end{smallmatrix}\bigr) \right |}$
$=\frac{\left |-3-9+2 \right |}{\sqrt{1+9+4}}$
$=\frac{5}{7}\sqrt{14}$

(iii) The graphic is as shown

So, point M is point B, moved along downwards along line l

Unit vector along line l =
$\frac{1}{\sqrt{14}}\bigl(\begin{smallmatrix} 1\\ -3\\ 2 \end{smallmatrix}\bigr)$

Hence,
$\vec{OM}=\bigl(\begin{smallmatrix} 1\\ -2\\ 3 \end{smallmatrix}\bigr)- \frac{5}{7}\sqrt{14}\left (\frac{1}{\sqrt{14}}\bigl(\begin{smallmatrix} 1\\ -3\\ 2 \end{smallmatrix}\bigr) \right )$
$=\begin{pmatrix} \frac{2}{7}\\ \frac{1}{7}\\ \frac{11}{7} \end{pmatrix}$

(iv) Let the point be
$\begin{pmatrix} 1\\ -2\\ 3 \end{pmatrix}+\begin{pmatrix} 1\\ -3\\ 2 \end{pmatrix}$

Meeting the plane z = 0 means 3 + 2λ = 0
λ = -1.5

Hence, substituting λ = -1.5, we get the point of intersection as
$\begin{pmatrix} 1-\frac{3}{2}\\ -2+\frac{9}{2}\\ 0 \end{pmatrix}=\begin{pmatrix} -\frac{1}{2}\\ \frac{5}{2}\\ 2 \end{pmatrix}$

v) Let
$\small \vec{OP}=\begin{pmatrix} 0\\ 1\\ 1 \end{pmatrix}+ k\begin{pmatrix} 1\\ -3\\ 2 \end{pmatrix}$ for some k since P is a point on l.

Note: So we need to find out what is the value of k.

$\small \vec{OP}=\begin{pmatrix} k\\ 1-3k\\ 1+2k \end{pmatrix}$

Since cos ∡POB = $\frac{2}{\sqrt{7}}$ , and
cos ∡POB = $\small \frac{\vec{OP}\cdot \vec{OB}}{\left |\vec{OP} \right |\left |\vec{OB} \right |}$
$\small =\frac{\begin{pmatrix} k\\ 1-3k\\ 1+2k \end{pmatrix}\cdot \begin{pmatrix} 1\\ -2\\ 3 \end{pmatrix}}{ \left |\begin{pmatrix} k\\ 1-3k\\ 1+2k \end{pmatrix} \right |\left |\begin{pmatrix} 1\\ -2\\ 3 \end{pmatrix} \right |}$
$\small \frac{2}{\sqrt{7}}=\frac{\begin{pmatrix} k\\ 1-3k\\ 1+2k \end{pmatrix}\cdot \begin{pmatrix} 1\\ -2\\ 3 \end{pmatrix}}{ \left |\begin{pmatrix} k\\ 1-3k\\ 1+2k \end{pmatrix} \right |\left |\begin{pmatrix} 1\\ -2\\ 3 \end{pmatrix} \right |}$
$\small \frac{2}{\sqrt{7}}=\frac{k-2+6k+3+6k}{\sqrt{14}\sqrt{k^{2}+(1-3k)^{2}+(1+2k)^{2}}}$
$\small \frac{2}{\sqrt{7}}=\frac{13k+1}{\sqrt{14}\sqrt{14k^{2}-2k+2}}$
$\small 13k+1=2\sqrt{2}\sqrt{14k^{2}-2k+2}$

Square both sides.

169k² + 26k + 1 = 112k² -16k + 16
57k² + 42k - 15 = 0
19k² + 14k - 5 = 0
(19k - 5)(k + 1) = 0
k = -1 or 5/19

Hence, the two points P are
(-1, 4, -1) or (5/19, 4/19, 29/19)

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A lvl H2 Maths: Vectors