Related Formula
The scalar projection of vector v$\vec{v}$ onto vector w$\vec{w}$ is calculated as:
Projection = v · w| w|$$\text{Projection} = \frac{\vec{v} \cdot \vec{w}}{|\vec{w}|}$$
Step 1: Calculate b$\vec{b}$
Compute the cross product using standard matrix expansion:
b = a × ( i - 2 k) = vmatrix i & j & k 3 & -1 & 2 1 & 0 & -2 vmatrix$$\vec{b} = \vec{a} \times (\hat{i} - 2\hat{k}) = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & -1 & 2 \\ 1 & 0 & -2 \end{vmatrix}$$
b = i(2 - 0) - j(-6 - 2) + k(0 - (-1)) = 2 i + 8 j + k$$\vec{b} = \hat{i}(2 - 0) - \hat{j}(-6 - 2) + \hat{k}(0 - (-1)) = 2\hat{i} + 8\hat{j} + \hat{k}$$
Step 2: Calculate c$\vec{c}$ and c - 2 j$\vec{c} - 2\hat{j}$
Perform the second cross product with unit vector k$\hat{k}$:
c = b × k = (2 i + 8 j + k) × k = 2( i × k) + 8( j × k) + 0$$\vec{c} = \vec{b} \times \hat{k} = (2\hat{i} + 8\hat{j} + \hat{k}) \times \hat{k} = 2(\hat{i} \times \hat{k}) + 8(\hat{j} \times \hat{k}) + \vec{0}$$
c = 2(- j) + 8( i) = 8 i - 2 j$$\vec{c} = 2(-\hat{j}) + 8(\hat{i}) = 8\hat{i} - 2\hat{j}$$
Subtract 2 j$2\hat{j}$:
c - 2 j = (8 i - 2 j) - 2 j = 8 i - 4 j$$\vec{c} - 2\hat{j} = (8\hat{i} - 2\hat{j}) - 2\hat{j} = 8\hat{i} - 4\hat{j}$$
Step 3: Compute the projection onto a$\vec{a}$
Using the dot product formula :
Projection = ( c - 2 j) · a| a| = 8, -4, 0 · 3, -1, 2 √(3² + (-1)² + 2²)$$\text{Projection} = \frac{(\vec{c} - 2\hat{j}) \cdot \vec{a}}{|\vec{a}|} = \frac{\langle 8, -4, 0 \rangle \cdot \langle 3, -1, 2 \rangle}{\sqrt{3^2 + (-1)^2 + 2^2}}$$
Projection = 24 + 4 + 0√(9 + 1 + 4) = 28√(14) = 2√(14)$$\text{Projection} = \frac{24 + 4 + 0}{\sqrt{9 + 1 + 4}} = \frac{28}{\sqrt{14}} = 2\sqrt{14}$$
Pattern Recognition
Keep cyclic unit cross products clear: i × k = - j$\hat{i} \times \hat{k} = -\hat{j}$ and j × k = i$\hat{j} \times \hat{k} = \hat{i}$. Missing a negative sign during basic cross multiplications ruins multi-step vector projections easily.
Chapter Mix
Class 12 Mathematics: Vector Algebra
