Related Formula
Length of projection of b on c = | b· c|| c|$$\text{Length of projection of } \vec{b} \text{ on } \vec{c} = \frac{|\vec{b}\cdot\vec{c}|}{|\vec{c}|}$$
Core Logic
First, find v = a × b$\vec{v} = \vec{a} \times \vec{b}$.
a = i - 2 j + 3 k$$\vec{a} = \hat{i} - 2\hat{j} + 3\hat{k}$$
b = 2 i + j - k$$\vec{b} = 2\hat{i} + \hat{j} - \hat{k}$$
v = vmatrix i & j & k 1 & -2 & 3 2 & 1 & -1 vmatrix = i(2-3) - j(-1-6) + k(1+4) = - i + 7 j + 5 k$$\vec{v} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -2 & 3 \\ 2 & 1 & -1 \end{vmatrix} = \hat{i}(2-3) - \hat{j}(-1-6) + \hat{k}(1+4) = -\hat{i} + 7\hat{j} + 5\hat{k}$$
Given v· c = 11$\vec{v}\cdot\vec{c} = 11$ and c = λ i + j + k$\vec{c} = \lambda\hat{i} + \hat{j} + \hat{k}$:
(- i + 7 j + 5 k) · (λ i + j + k) = 11$$(-\hat{i} + 7\hat{j} + 5\hat{k}) \cdot (\lambda\hat{i} + \hat{j} + \hat{k}) = 11$$
-λ + 7 + 5 = 11 λ = 1$$-\lambda + 7 + 5 = 11 \implies \lambda = 1$$
Step 1: Projection Calculation
Now, c = i + j + k$\vec{c} = \hat{i} + \hat{j} + \hat{k}$.
The length of the projection of b$\vec{b}$ on c$\vec{c}$ is p$p$:
p = | (2 i + j - k) · ( i + j + k)√(1² + 1² + 1²) | = 2(1) + 1(1) - 1(1)√(3) = 2√(3)$$p = \left| \frac{(2\hat{i} + \hat{j} - \hat{k}) \cdot (\hat{i} + \hat{j} + \hat{k})}{\sqrt{1^2 + 1^2 + 1^2}} \right| = \frac{2(1) + 1(1) - 1(1)}{\sqrt{3}} = \frac{2}{\sqrt{3}}$$
Calculate 9p²$9p^2$:
9p² = 9((4)/(3)) = 12$$9p^2 = 9\left(\frac{4}{3}\right) = 12$$
Pattern Recognition
Standard sequence: compute cross product to find normal vector v$\vec{v}$, take dot product to deduce missing parameter λ$\lambda$, then substitute into scalar projection formula.
Chapter Mix
Class 12 Maths: Vector Algebra