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Geometry and Trigonometry - Vector equations (HL)

Grade 11IB_AI

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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A line in 3D space is uniquely defined by a position vector a\mathbf{a} (a fixed point on the line) and a direction vector b\mathbf{b} (indicating the line's orientation). The general equation is r=a+λb\mathbf{r} = \mathbf{a} + \lambda \mathbf{b}, where λ\lambda is a scalar parameter.

Vector representation of a line showing position vector a and direction vector b from the origin O.
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Two lines in 3D space can be parallel, intersecting, or skew. Skew lines are non-parallel lines that do not intersect because they lie in different planes.

Diagram showing skew lines in two parallel planes.
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The angle θ\theta between two lines is determined solely by the dot product of their direction vectors b1\mathbf{b_1} and b2\mathbf{b_2}. If b1⋅b2=0\mathbf{b_1} \cdot \mathbf{b_2} = 0, the lines are perpendicular.

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The shortest distance from a point to a line can be found by creating a vector from the point to a general point on the line and ensuring it is perpendicular to the direction vector.

📐Formulae

r=a+λb\mathbf{r} = \mathbf{a} + \lambda \mathbf{b}

(xyz)=\begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} x_0 \ y_0 \ z_0 \end{pmatrix}+λ(lmn) + \lambda \begin{pmatrix} l \\ m \\ n \end{pmatrix}

cos⁡θ=∣b1⋅b2∣∣b1∣∣b2∣\cos \theta = \frac{|\mathbf{b_1} \cdot \mathbf{b_2}|}{|\mathbf{b_1}| |\mathbf{b_2}|}

a⋅b=a1b1+a2b2+a3b3\mathbf{a} \cdot \mathbf{b} = a_1 b_1 + a_2 b_2 + a_3 b_3

∣v∣=v12+v22+v32|\mathbf{v}| = \sqrt{v_1^2 + v_2^2 + v_3^2}

💡Examples

Problem 1:

Find the vector equation of the line LL passing through the points A(2,−1,4)A(2, -1, 4) and B(5,3,−2)B(5, 3, -2).

Solution:

First, find the direction vector b=AB⃗\mathbf{b} = \vec{AB}: AB⃗=\vec{AB} = \begin{pmatrix} 5 - 2 \ 3 - (-1) \ -2 - 4 \end{pmatrix}==\begin{pmatrix} 3 \ 4 \ -6 \end{pmatrix}Using point $A$ as the position vector $\mathbf{a}$:\mathbf{r} = \begin{pmatrix} 2 \ -1 \ 4 \end{pmatrix}+λ(34−6) + \lambda \begin{pmatrix} 3 \\ 4 \\ -6 \end{pmatrix}

Explanation:

To find the vector equation, we need a point on the line (position vector) and the vector connecting two points (direction vector). Subtract coordinates of the start point from the end point to get the direction.

Problem 2:

Determine if the lines L_1: \mathbf{r} = $$\begin{pmatrix} 1 \\ 2 \\ 3 \end{pmatrix}$$ + \lambda $$\begin{pmatrix} 1 \\ 0 \\ -1 \end{pmatrix}$$ and L_2: \mathbf{r} = \begin{pmatrix} 2 \ 2 \ 2 \end{pmatrix}$$ + \mu \begin{pmatrix} 0 \ 1 \ 1 \end{pmatrix}$ intersect.

Solution:

Equate the components of L1L_1 and L2L_2:

  1. 1+λ=2⇒λ=11 + \lambda = 2 \Rightarrow \lambda = 1
  2. 2+0λ=2+μ⇒μ=02 + 0\lambda = 2 + \mu \Rightarrow \mu = 0
  3. 3−λ=2+μ3 - \lambda = 2 + \mu Substitute λ=1\lambda = 1 and μ=0\mu = 0 into the third equation: 3−1=2+0⇒2=23 - 1 = 2 + 0 \Rightarrow 2 = 2 Since the values are consistent, they intersect. The point of intersection is found by substituting λ=1\lambda = 1 into L1L_1: r=\mathbf{r} = \begin{pmatrix} 1+1 \ 2+0 \ 3-1 \end{pmatrix}=(222) = \begin{pmatrix} 2 \\ 2 \\ 2 \end{pmatrix}

Explanation:

To check for intersection, create a system of three linear equations based on the x,y,zx, y, z components. If a consistent pair of parameters (λ,μ)(\lambda, \mu) satisfies all three equations, the lines intersect.

Problem 3:

Find the acute angle between the lines with direction vectors d1=(122)\mathbf{d_1} = \begin{pmatrix} 1 \\ 2 \\ 2 \end{pmatrix} and d2=(304)\mathbf{d_2} = \begin{pmatrix} 3 \\ 0 \\ 4 \end{pmatrix}.

Solution:

Calculate the dot product: d1⋅d2=(1)(3)+(2)(0)+(2)(4)=3+0+8=11\mathbf{d_1} \cdot \mathbf{d_2} = (1)(3) + (2)(0) + (2)(4) = 3 + 0 + 8 = 11. Calculate magnitudes: ∣d1∣=12+22+22=9=3|\mathbf{d_1}| = \sqrt{1^2 + 2^2 + 2^2} = \sqrt{9} = 3 ∣d2∣=32+02+42=25=5|\mathbf{d_2}| = \sqrt{3^2 + 0^2 + 4^2} = \sqrt{25} = 5 Apply the formula: cos⁡θ=113×5=1115\cos \theta = \frac{11}{3 \times 5} = \frac{11}{15} θ=arccos⁡(1115)≈42.8∘\theta = \arccos\left(\frac{11}{15}\right) \approx 42.8^\circ

Explanation:

The angle between two lines is defined as the angle between their direction vectors. We use the cosine rule for dot products to solve for θ\theta.

Problem 4:

Calculate the coordinates of the point of intersection between the lines L1:r=(101)+λ(21−1)L_1: \mathbf{r} = \begin{pmatrix} 1 \\ 0 \\ 1 \end{pmatrix} + \lambda \begin{pmatrix} 2 \\ 1 \\ -1 \end{pmatrix} and L2:r=(234)+μ(−1−2−3)L_2: \mathbf{r} = \begin{pmatrix} 2 \\ 3 \\ 4 \end{pmatrix} + \mu \begin{pmatrix} -1 \\ -2 \\ -3 \end{pmatrix}.

Two non-intersecting lines L1 and L2 projected on a 2D plane to illustrate the search for an intersection.

Solution:

Set the components equal:

  1. 1+2λ=2−μ⇒2λ+μ=11 + 2\lambda = 2 - \mu \Rightarrow 2\lambda + \mu = 1
  2. 0+λ=3−2μ⇒λ+2μ=30 + \lambda = 3 - 2\mu \Rightarrow \lambda + 2\mu = 3
  3. 1−λ=4−3μ⇒−λ+3μ=31 - \lambda = 4 - 3\mu \Rightarrow -\lambda + 3\mu = 3

Solving (1) and (2): Multiply (1) by 2: 4λ+2μ=24\lambda + 2\mu = 2 Subtract (2): 3λ=−1⇒λ=−133\lambda = -1 \Rightarrow \lambda = -\frac{1}{3} Substitute λ\lambda into (1): 2(−13)+μ=1⇒μ=1+23=532(-\frac{1}{3}) + \mu = 1 \Rightarrow \mu = 1 + \frac{2}{3} = \frac{5}{3}

Check in (3): −−13+3(53)=13+5=163≠3-\frac{-1}{3} + 3(\frac{5}{3}) = \frac{1}{3} + 5 = \frac{16}{3} \neq 3. Since the values do not satisfy the third equation, the lines do not intersect (they are skew).

Explanation:

To find an intersection, we equate the x,y,x, y, and zz components to create a system of equations. We solve for the parameters using two equations and verify with the third. If it doesn't match, they are skew.

Problem 5:

Find the vector equation of the line that passes through the point P(1,2,3)P(1, 2, 3) and is parallel to the line r=(456)+t(−102)\mathbf{r} = \begin{pmatrix} 4 \\ 5 \\ 6 \end{pmatrix} + t \begin{pmatrix} -1 \\ 0 \\ 2 \end{pmatrix}.

Diagram showing two parallel lines, one passing through point P, both sharing the same direction.

Solution:

Parallel lines share the same direction vector. The direction vector of the given line is b=(−102)\mathbf{b} = \begin{pmatrix} -1 \\ 0 \\ 2 \end{pmatrix}. The position vector of the point PP is a=(123)\mathbf{a} = \begin{pmatrix} 1 \\ 2 \\ 3 \end{pmatrix}. The equation of the new line is: r=(123)+λ(−102)\mathbf{r} = \begin{pmatrix} 1 \\ 2 \\ 3 \end{pmatrix} + \lambda \begin{pmatrix} -1 \\ 0 \\ 2 \end{pmatrix}

Explanation:

Since the lines are parallel, we use the direction vector from the known line and the coordinates of the given point as the starting position vector.