Related Formula
Any vector a$\vec{a}$ can be written as the sum of its component parallel to b$\vec{b}$ (along b$\vec{b}$) and its component perpendicular to b$\vec{b}$:
a = a∥ + a⊥$$\vec{a} = \vec{a}_{\parallel} + \vec{a}_{\perp}$$
Magnitude squared:
| a|² = α² + β² + γ²$$|\vec{a}|^2 = \alpha^2 + \beta^2 + \gamma^2$$
Core Logic
Given:
a∥ = (16)/(11)(3 i+ j- k)$$\vec{a}_{\parallel} = \frac{16}{11}(3\hat{i}+\hat{j}-\hat{k})$$
a⊥ = (1)/(11)(-4 i-5 j-17 k)$$\vec{a}_{\perp} = \frac{1}{11}(-4\hat{i}-5\hat{j}-17\hat{k})$$
Step 1: Reconstruct Vector a
Add both components to find a$\vec{a}$:
a = (16)/(11)(3 i+ j- k) + (1)/(11)(-4 i-5 j-17 k)$$\vec{a} = \frac{16}{11}(3\hat{i}+\hat{j}-\hat{k}) + \frac{1}{11}(-4\hat{i}-5\hat{j}-17\hat{k})$$
a = (1)/(11) [ (48 - 4) i + (16 - 5) j + (-16 - 17) k ]$$\vec{a} = \frac{1}{11} \left[ (48 - 4)\hat{i} + (16 - 5)\hat{j} + (-16 - 17)\hat{k} \right]$$
a = (1)/(11) [ 44 i + 11 j - 33 k ] = 4 i + j - 3 k$$\vec{a} = \frac{1}{11} \left[ 44\hat{i} + 11\hat{j} - 33\hat{k} \right] = 4\hat{i} + \hat{j} - 3\hat{k}$$
Therefore, α = 4$\alpha = 4$, β = 1$\beta = 1$, γ = -3$\gamma = -3$.
Step 2: Calculate Sum of Squares
α² + β² + γ² = 4² + 1² + (-3)² = 16 + 1 + 9 = 26$$\alpha^2 + \beta^2 + \gamma^2 = 4^2 + 1^2 + (-3)^2 = 16 + 1 + 9 = 26$$
Pattern Recognition
Since parallel and perpendicular components are orthogonal vectors, you can also use directly | a|² = | a∥|² + | a⊥|²$|\vec{a}|^2 = |\vec{a}_{\parallel}|^2 + |\vec{a}_{\perp}|^2$ to save algebra step: | a∥|² = (16²)/(11²)(9+1+1) = (256)/(11)$|\vec{a}_{\parallel}|^2 = \frac{16^2}{11^2}(9+1+1) = \frac{256}{11}$, | a⊥|² = (1)/(11²)(16+25+289) = (330)/(121) = (30)/(11)$|\vec{a}_{\perp}|^2 = \frac{1}{11^2}(16+25+289) = \frac{330}{121} = \frac{30}{11}$. Total = (286)/(11) = 26$\frac{286}{11} = 26$.
Chapter Mix
Class 12 Mathematics: Vector Algebra
