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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

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

The scalar projection (also called the component) of a\vec{a} on b\vec{b} is a real number given by acosθ|\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 b\vec{b}; if obtuse, it is negative and points in the opposite direction.

The dot product is the fundamental tool for calculating projections because ab=abcosθ\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.

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 b\vec{b}. Visually, this is the actual vector arrow that lies along the line of b\vec{b} starting from the same origin as a\vec{a}.

When two vectors are perpendicular (orthogonal), the angle θ=90\theta = 90^\circ and cos90=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.

The projection of a vector r=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 r\vec{r} on the xx-axis is the distance from the origin to the point where a perpendicular line dropped from the tip of r\vec{r} meets the xx-axis.

📐Formulae

Scalar Projection of a\vec{a} on b=abb\vec{b} = \frac{\vec{a} \cdot \vec{b}}{|\vec{b}|}

Vector Projection of a\vec{a} on b=(abb2)b\vec{b} = \left( \frac{\vec{a} \cdot \vec{b}}{|\vec{b}|^2} \right) \vec{b}

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

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

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

💡Examples

Problem 1:

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

Solution:

  1. First, calculate the dot product ab\vec{a} \cdot \vec{b}: ab=(2)(1)+(1)(2)+(1)(2)=22+2=2\vec{a} \cdot \vec{b} = (2)(1) + (-1)(2) + (1)(2) = 2 - 2 + 2 = 2
  2. Next, find the magnitude of b\vec{b}: b=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=abb=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 a\vec{a} aligns with b\vec{b} by dividing their dot product by the length of the target vector b\vec{b}.

Problem 2:

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

Solution:

  1. Calculate pq\vec{p} \cdot \vec{q}: pq=(1)(4)+(3)(3)=49=5\vec{p} \cdot \vec{q} = (1)(4) + (3)(-3) = 4 - 9 = -5
  2. Calculate q2|\vec{q}|^2: q2=42+(3)2=16+9=25|\vec{q}|^2 = 4^2 + (-3)^2 = 16 + 9 = 25
  3. Use the vector projection formula: v=(pqq2)q\vec{v} = \left( \frac{\vec{p} \cdot \vec{q}}{|\vec{q}|^2} \right) \vec{q} v=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}) v=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 q\vec{q}.