Why do I enter s and t?

The coefficients in \overrightarrow{\mathrm{OP}} = s\vec{a} + t\vec{b} determine the vector uniquely. You can type "s = 1/3, t = 1/2" or "(1/3, 1/2)".

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Vectors in the Plane: Practice Problems

Plane vectors combine component calculations with geometry through position vectors. Master the dot product and the idea of comparing coefficients.

Basic, Standard: Grade 11 · Term 3 / Advanced: Grade 12 · Exam prep

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Components, magnitude and dot product

Formulas

|\vec{a}| = \sqrt{a_1^2 + a_2^2},\qquad \vec{a} \cdot \vec{b} = a_1b_1 + a_2b_2 = |\vec{a}||\vec{b}|\cos\theta

Find angles from $\cos\theta = \dfrac{\vec{a} \cdot \vec{b}}{|\vec{a}||\vec{b}|}$ . For magnitudes, square and expand: $|\vec{a} - \vec{b}|^2 = |\vec{a}|^2 - 2\vec{a} \cdot \vec{b} + |\vec{b}|^2$ .

Parallel and perpendicular

$\vec{a} \parallel \vec{b} \iff \vec{a} = k\vec{b}$ for a real $k$ ; $\vec{a} \perp \vec{b} \iff \vec{a} \cdot \vec{b} = 0$ (for nonzero vectors).

Position vectors

Section formula

\text{internal } m : n:\ \frac{n\vec{a} + m\vec{b}}{m + n},\qquad \text{external } m : n:\ \frac{-n\vec{a} + m\vec{b}}{m - n},\qquad \text{centroid: } \frac{\vec{a} + \vec{b} + \vec{c}}{3}

For an angle bisector, use $\mathrm{AD} : \mathrm{DB} = \mathrm{OA} : \mathrm{OB}$ and then the section formula.

Intersection points

To find the intersection $\mathrm{P}$ of two segments, write $\overrightarrow{\mathrm{OP}}$ in two ways and compare coefficients. This is valid because $\vec{a}$ and $\vec{b}$ are linearly independent (nonzero and not parallel).

Areas and area ratios

Area of a triangle

S = \frac{1}{2}\sqrt{|\vec{a}|^2|\vec{b}|^2 - (\vec{a} \cdot \vec{b})^2} = \frac{1}{2}|a_1b_2 - a_2b_1|

If $\alpha\overrightarrow{\mathrm{PA}} + \beta\overrightarrow{\mathrm{PB}} + \gamma\overrightarrow{\mathrm{PC}} = \vec{0}$ with $\alpha, \beta, \gamma > 0$ , then $\triangle \mathrm{PBC} : \triangle \mathrm{PCA} : \triangle \mathrm{PAB} = \alpha : \beta : \gamma$ .

Q1BasicGr 11 · Term 3Vectors in the plane

Given $\mathrm{A}(0,\ 2)$ , $\mathrm{B}(5,\ 0)$ and $\mathrm{C}(7,\ 1)$ , find the point $\mathrm{D}$ such that $\mathrm{ABCD}$ is a parallelogram.

Answer

\mathrm{D}(2,\ 3)

Solution

Since it is a parallelogram, $\overrightarrow{\mathrm{AD}} = \overrightarrow{\mathrm{BC}}$ .
$\overrightarrow{\mathrm{BC}} = (2,\ 1)$
So
$\mathrm{D} = (0,\ 2) + (2,\ 1) = (2,\ 3)$

Q2StandardGr 11 · Term 3Vectors in the plane

In $\triangle \mathrm{OAB}$ , let $\overrightarrow{\mathrm{OA}} = \vec{a}$ and $\overrightarrow{\mathrm{OB}} = \vec{b}$ . Let $\mathrm{P}$ divide $\mathrm{AB}$ internally in the ratio $3 : 5$ , and $\mathrm{Q}$ divide it externally in the ratio $3 : 5$ . Express $\overrightarrow{\mathrm{OP}}$ and $\overrightarrow{\mathrm{OQ}}$ in terms of $\vec{a}$ and $\vec{b}$ .

(1) $\overrightarrow{\mathrm{OP}}$ (Write $\overrightarrow{\mathrm{OP}} = s\vec{a} + t\vec{b}$ and give $s$ and $t$ .)
(2) $\overrightarrow{\mathrm{OQ}}$ (Write $\overrightarrow{\mathrm{OQ}} = s\vec{a} + t\vec{b}$ and give $s$ and $t$ .)

Answer

(1) $\overrightarrow{\mathrm{OP}} = \frac{5}{8}\vec{a} + \frac{3}{8}\vec{b}$
(2) $\overrightarrow{\mathrm{OQ}} = \frac{5}{2}\vec{a} - \frac{3}{2}\vec{b}$

Solution

By the section formula (internal),
$\overrightarrow{\mathrm{OP}} = \frac{5\vec{a} + 3\vec{b}}{3 + 5} = \frac{5}{8}\vec{a} + \frac{3}{8}\vec{b}$
By the section formula (external),
$\overrightarrow{\mathrm{OQ}} = \frac{-5\vec{a} + 3\vec{b}}{3 - 5} = \frac{5}{2}\vec{a} - \frac{3}{2}\vec{b}$

Q3AdvancedExam prepVectors in the plane

In $\triangle \mathrm{OAB}$ , let $\overrightarrow{\mathrm{OA}} = \vec{a}$ and $\overrightarrow{\mathrm{OB}} = \vec{b}$ , with $|\vec{a}| = 3$ , $|\vec{b}| = 3$ and $\vec{a} \cdot \vec{b} = -5$ . Let $\mathrm{P}$ be the circumcenter of $\triangle \mathrm{OAB}$ . Express $\overrightarrow{\mathrm{OP}}$ in terms of $\vec{a}$ and $\vec{b}$ . (Write $\overrightarrow{\mathrm{OP}} = s\vec{a} + t\vec{b}$ and give $s$ and $t$ .)

Answer

\overrightarrow{\mathrm{OP}} = \frac{9}{8}\vec{a} + \frac{9}{8}\vec{b}

Solution

Let $\overrightarrow{\mathrm{OP}} = s\vec{a} + t\vec{b}$ .
$\overrightarrow{\mathrm{OP}} \cdot \vec{a} = \frac{1}{2}|\vec{a}|^2,\qquad \overrightarrow{\mathrm{OP}} \cdot \vec{b} = \frac{1}{2}|\vec{b}|^2$
Substituting the given values,
$9s - 5t = \frac{9}{2},\qquad -5s + 9t = \frac{9}{2}$
Solving,
$\overrightarrow{\mathrm{OP}} = \frac{9}{8}\vec{a} + \frac{9}{8}\vec{b}$

Q1BasicGr 11 · Term 3Vectors in the plane

For $\vec{a} = (2,\ 0)$ and $\vec{b} = (-2,\ 2)$ , find the dot product $\vec{a} \cdot \vec{b}$ and the angle $\theta$ between them. Give $\theta$ in degrees with $0^\circ \leqq \theta \leqq 180^\circ$ .

(1) $\vec{a} \cdot \vec{b}$
(2) $\theta$

Q2StandardGr 11 · Term 3Vectors in the plane

If $|\overrightarrow{\mathrm{OA}}| = 6$ , $|\overrightarrow{\mathrm{OB}}| = 3$ and $\overrightarrow{\mathrm{OA}} \cdot \overrightarrow{\mathrm{OB}} = 14$ , find the area of $\triangle \mathrm{OAB}$ .

Q3AdvancedExam prepVectors in the plane

In $\triangle \mathrm{OAB}$ , let $\overrightarrow{\mathrm{OA}} = \vec{a}$ and $\overrightarrow{\mathrm{OB}} = \vec{b}$ , with $|\vec{a}| = 4$ , $|\vec{b}| = 3$ and $\vec{a} \cdot \vec{b} = -3$ . Let $\mathrm{H}$ be the orthocenter of $\triangle \mathrm{OAB}$ . Express $\overrightarrow{\mathrm{OH}}$ in terms of $\vec{a}$ and $\vec{b}$ . (Write $\overrightarrow{\mathrm{OH}} = s\vec{a} + t\vec{b}$ and give $s$ and $t$ .)

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