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Vectors - Projection of a Vector on a Line

Grade 12ICSE

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

🔑Concepts

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The projection of a vector a⃗\vec{a} on another vector b⃗\vec{b} is essentially the 'shadow' of vector a⃗\vec{a} cast onto the line containing vector b⃗\vec{b}. Visually, if you imagine a light source positioned perpendicular to b⃗\vec{b}, the length of the segment on b⃗\vec{b} covered by the shadow of a⃗\vec{a} represents the magnitude of the projection.

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The scalar projection (also called the component) of a⃗\vec{a} on b⃗\vec{b} is a real number given by ∣a⃗∣cos⁡θ|\vec{a}| \cos \theta, where θ\theta is the angle between the two vectors. If the angle is acute, the projection is positive and points in the direction of b⃗\vec{b}; if obtuse, it is negative and points in the opposite direction.

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The dot product is the fundamental tool for calculating projections because a⃗⋅b⃗=∣a⃗∣∣b⃗∣cos⁡θ\vec{a} \cdot \vec{b} = |\vec{a}||\vec{b}| \cos \theta. This relationship allows us to find the scalar component without directly calculating the angle θ\theta.

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A vector projection differs from a scalar projection because it is a vector quantity. It is obtained by multiplying the scalar projection by the unit vector in the direction of b⃗\vec{b}. Visually, this is the actual vector arrow that lies along the line of b⃗\vec{b} starting from the same origin as a⃗\vec{a}.

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When two vectors are perpendicular (orthogonal), the angle θ=90∘\theta = 90^\circ and cos⁡90∘=0\cos 90^\circ = 0. Consequently, the projection of one onto the other is zero. Geometrically, a vertical object cast no shadow on a horizontal line when the light source is directly above it.

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The projection of a vector r⃗=xi^+yj^+zk^\vec{r} = x\hat{i} + y\hat{j} + z\hat{k} on the coordinate axes (x,y,zx, y, z) are simply its components xx, yy, and zz. For example, the projection of r⃗\vec{r} on the xx-axis is the distance from the origin to the point where a perpendicular line dropped from the tip of r⃗\vec{r} meets the xx-axis.

📐Formulae

Scalar Projection of a⃗\vec{a} on b⃗=a⃗⋅b⃗∣b⃗∣\vec{b} = \frac{\vec{a} \cdot \vec{b}}{|\vec{b}|}

Vector Projection of a⃗\vec{a} on b⃗=(a⃗⋅b⃗∣b⃗∣2)b⃗\vec{b} = \left( \frac{\vec{a} \cdot \vec{b}}{|\vec{b}|^2} \right) \vec{b}

Dot Product: a⃗⋅b⃗=a1b1+a2b2+a3b3\vec{a} \cdot \vec{b} = a_1b_1 + a_2b_2 + a_3b_3

Magnitude: ∣b⃗∣=b12+b22+b32|\vec{b}| = \sqrt{b_1^2 + b_2^2 + b_3^2}

Unit Vector: b^=b⃗∣b⃗∣\hat{b} = \frac{\vec{b}}{|\vec{b}|}

💡Examples

Problem 1:

Find the scalar projection of the vector a⃗=2i^−j^+k^\vec{a} = 2\hat{i} - \hat{j} + \hat{k} on the vector b⃗=i^+2j^+2k^\vec{b} = \hat{i} + 2\hat{j} + 2\hat{k}.

Solution:

  1. First, calculate the dot product a⃗⋅b⃗\vec{a} \cdot \vec{b}: a⃗⋅b⃗=(2)(1)+(−1)(2)+(1)(2)=2−2+2=2\vec{a} \cdot \vec{b} = (2)(1) + (-1)(2) + (1)(2) = 2 - 2 + 2 = 2
  2. Next, find the magnitude of b⃗\vec{b}: ∣b⃗∣=12+22+22=1+4+4=9=3|\vec{b}| = \sqrt{1^2 + 2^2 + 2^2} = \sqrt{1 + 4 + 4} = \sqrt{9} = 3
  3. Use the formula for scalar projection: Projection=a⃗⋅b⃗∣b⃗∣=23\text{Projection} = \frac{\vec{a} \cdot \vec{b}}{|\vec{b}|} = \frac{2}{3}

Explanation:

To find the scalar projection, we determine how much of a⃗\vec{a} aligns with b⃗\vec{b} by dividing their dot product by the length of the target vector b⃗\vec{b}.

Problem 2:

Find the vector projection of p⃗=i^+3j^\vec{p} = \hat{i} + 3\hat{j} on q⃗=4i^−3j^\vec{q} = 4\hat{i} - 3\hat{j}.

Solution:

  1. Calculate p⃗⋅q⃗\vec{p} \cdot \vec{q}: p⃗⋅q⃗=(1)(4)+(3)(−3)=4−9=−5\vec{p} \cdot \vec{q} = (1)(4) + (3)(-3) = 4 - 9 = -5
  2. Calculate ∣q⃗∣2|\vec{q}|^2: ∣q⃗∣2=42+(−3)2=16+9=25|\vec{q}|^2 = 4^2 + (-3)^2 = 16 + 9 = 25
  3. Use the vector projection formula: v⃗=(p⃗⋅q⃗∣q⃗∣2)q⃗\vec{v} = \left( \frac{\vec{p} \cdot \vec{q}}{|\vec{q}|^2} \right) \vec{q} v⃗=−525(4i^−3j^)=−15(4i^−3j^)\vec{v} = \frac{-5}{25}(4\hat{i} - 3\hat{j}) = -\frac{1}{5}(4\hat{i} - 3\hat{j}) v⃗=−0.8i^+0.6j^\vec{v} = -0.8\hat{i} + 0.6\hat{j}

Explanation:

The vector projection results in a vector. Since the dot product is negative, the resulting vector points in the opposite direction of q⃗\vec{q}.