Math C

Vectors in Space: Practice Problems

Space vectors work just like plane vectors with one more component. The new idea is the condition for a point to lie in a plane.

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

BasicStandardAdvancedMix

0Make a worksheet on this topic

Components and dot product

Formulas

|\vec{a}| = \sqrt{a_1^2 + a_2^2 + a_3^2},\qquad \vec{a} \cdot \vec{b} = a_1b_1 + a_2b_2 + a_3b_3

A vector perpendicular to both $\vec{a}$ and $\vec{b}$ is found by writing it as $(x,\ y,\ z)$ and setting both dot products to $0$ .

Points in a plane

P lies in plane ABC

\overrightarrow{\mathrm{AP}} = s\overrightarrow{\mathrm{AB}} + t\overrightarrow{\mathrm{AC}}

\overrightarrow{\mathrm{OP}} = s\overrightarrow{\mathrm{OA}} + t\overrightarrow{\mathrm{OB}} + u\overrightarrow{\mathrm{OC}},\quad s + t + u = 1

The second form ("coefficients add up to 1") is the quickest way to find where a line meets a plane.

Line–plane intersections and perpendiculars

For the intersection $\mathrm{Q}$ of line $\mathrm{OP}$ with plane $\mathrm{ABC}$ , write $\overrightarrow{\mathrm{OQ}} = k\overrightarrow{\mathrm{OP}}$ and choose $k$ so that the coefficients add up to $1$ .

For the foot $\mathrm{H}$ of the perpendicular from a point to a plane, combine "H is in the plane" with " $\overrightarrow{\mathrm{OH}}$ is perpendicular to two vectors in the plane". The volume of a tetrahedron is base area × height ÷ 3.

Q1BasicGr 11 · Term 3Vectors in space

The vectors $\vec{a} = (1,\ x,\ -3)$ and $\vec{b} = (2,\ 1,\ -2)$ are perpendicular. Find $x$ .

Answer

x = -8

Solution

The dot product must be 0.
$1 \cdot 2 + x \cdot 1 + (-3) \cdot (-2) = 0 \quad\Longrightarrow\quad x + 8 = 0$
Solving,
$x = -8$

Q2StandardGr 11 · Term 3Vectors in space

Find the vector of magnitude $14$ that is perpendicular to both $\vec{a} = (2,\ 0,\ -3)$ and $\vec{b} = (2,\ 1,\ -6)$ and has a positive $x$ -component.

Answer

(6,\ 12,\ 4)

Solution

Let $\vec{p} = (x,\ y,\ z)$ . Then
$\vec{a} \cdot \vec{p} = 2x - 3z = 0,\qquad \vec{b} \cdot \vec{p} = 2x + y - 6z = 0$
Expressing $y$ and $z$ in terms of $x$ ,
$y = 2x,\qquad z = \frac{2}{3}x$
Since the magnitude is $14$ ,
$x^2 + \left(2x\right)^2 + \left(\frac{2}{3}x\right)^2 = 196$
Since $x > 0$ , $x = 6$ , so
$(6,\ 12,\ 4)$

Q3AdvancedExam prepVectors in space

Given $\mathrm{O}(0,\ 0,\ 0)$ , $\mathrm{A}(2,\ 0,\ 0)$ , $\mathrm{B}(0,\ 2,\ 0)$ , $\mathrm{C}(0,\ 0,\ 2)$ and $\mathrm{P}(1,\ 2,\ 1)$ , find the point $\mathrm{Q}$ where the line $\mathrm{OP}$ meets the plane $\mathrm{ABC}$ .

Answer

\mathrm{Q}\left(\frac{1}{2},\ 1,\ \frac{1}{2}\right)

Solution

Q is on line OP, so $\overrightarrow{\mathrm{OQ}} = k\overrightarrow{\mathrm{OP}}$ . Since Q is also in plane ABC,
$\overrightarrow{\mathrm{OQ}} = s\overrightarrow{\mathrm{OA}} + t\overrightarrow{\mathrm{OB}} + u\overrightarrow{\mathrm{OC}},\quad s + t + u = 1$
Comparing components,
$s = \frac{1}{2}k,\quad t = k,\quad u = \frac{1}{2}k$
From $s + t + u = 1$ ,
$k = \frac{1}{2},\qquad \mathrm{Q}\left(\frac{1}{2},\ 1,\ \frac{1}{2}\right)$

Q1BasicGr 11 · Term 3Vectors in space

The vectors $\vec{a} = (-3,\ x,\ 3)$ and $\vec{b} = (-1,\ 2,\ -3)$ are perpendicular. Find $x$ .

Q2StandardGr 11 · Term 3Vectors in space

Find the vector of magnitude $9$ that is perpendicular to both $\vec{a} = (4,\ -7,\ 0)$ and $\vec{b} = (8,\ -7,\ -7)$ and has a positive $x$ -component.

Q3AdvancedExam prepVectors in space

In the tetrahedron $\mathrm{OABC}$ , let $\overrightarrow{\mathrm{OA}} = \vec{a}$ , $\overrightarrow{\mathrm{OB}} = \vec{b}$ and $\overrightarrow{\mathrm{OC}} = \vec{c}$ . Let $\mathrm{L}$ be the point dividing $\mathrm{OA}$ internally in the ratio $1 : 2$ , $\mathrm{M}$ be the midpoint of $\mathrm{BC}$ , and $\mathrm{P}$ be the midpoint of $\mathrm{LM}$ . The line $\mathrm{OP}$ meets the plane $\mathrm{ABC}$ at $\mathrm{Q}$ . Express $\overrightarrow{\mathrm{OQ}}$ in terms of $\vec{a}, \vec{b}, \vec{c}$ . (Write $\overrightarrow{\mathrm{OQ}} = s\vec{a} + t\vec{b} + u\vec{c}$ and give $s, t, u$ .)

Teleport

Main features

On this page

Topics

Display

Vectors in Space: Practice Problems

Components and dot product

Points in a plane

Line–plane intersections and perpendiculars

Hint

Answer

Solution

Hint

Answer

Solution

Hint

Answer

Solution

Hint

Answer

Solution

Hint

Answer

Solution

Hint

Answer

Solution

Math problem generator

Components and dot product

Points in a plane

Line–plane intersections and perpendiculars

Worked examples

Hint

Answer

Solution

Hint

Answer

Solution

Hint

Answer

Solution

Practice problems

Hint

Answer

Solution

Hint

Answer

Solution

Hint

Answer

Solution

Related topics

Vectors in the plane