Related Formula
x · ( y × z) = [ x y z] = ( x × y) · z$$\vec{x} \cdot (\vec{y} \times \vec{z}) = [\vec{x} \ \vec{y} \ \vec{z}] = (\vec{x} \times \vec{y}) \cdot \vec{z}$$
Core Logic
The expression to evaluate is:
E = a · (( c × b) - b - c)$$E = \vec{a} \cdot ((\vec{c} \times \vec{b}) - \vec{b} - \vec{c})$$
Distributing the dot product over the terms gives:
E = a · ( c × b) - a · b - a · c$$E = \vec{a} \cdot (\vec{c} \times \vec{b}) - \vec{a} \cdot \vec{b} - \vec{a} \cdot \vec{c}$$
The first term is a scalar triple product:
a · ( c × b) = ( a × c) · b$$\vec{a} \cdot (\vec{c} \times \vec{b}) = (\vec{a} \times \vec{c}) \cdot \vec{b}$$
We are given that a × c = b$\vec{a} \times \vec{c} = \vec{b}$. Substitute this in:
( b) · b = | b|²$$(\vec{b}) \cdot \vec{b} = |\vec{b}|^2$$
Given b = 3 i - 3 j + 3 k$\vec{b} = 3\hat{i} - 3\hat{j} + 3\hat{k}$, its magnitude squared is:
| b|² = 3² + (-3)² + 3² = 9 + 9 + 9 = 27$$|\vec{b}|^2 = 3^2 + (-3)^2 + 3^2 = 9 + 9 + 9 = 27$$
Step 2: Evaluating the remaining Dot Products
For the second term, calculate a · b$\vec{a} \cdot \vec{b}$:
a = 1 i + 2 j + 1 k$$\vec{a} = 1\hat{i} + 2\hat{j} + 1\hat{k}$$
b = 3 i - 3 j + 3 k$$\vec{b} = 3\hat{i} - 3\hat{j} + 3\hat{k}$$
a · b = (1)(3) + (2)(-3) + (1)(3) = 3 - 6 + 3 = 0$$\vec{a} \cdot \vec{b} = (1)(3) + (2)(-3) + (1)(3) = 3 - 6 + 3 = 0$$
For the third term, we are explicitly given:
a · c = 3$$\vec{a} \cdot \vec{c} = 3$$
Step 3: Final Output Calculation
Substitute all individual values back into the expanded expression:
E = 27 - 0 - 3 = 24$$E = 27 - 0 - 3 = 24$$
Pattern Recognition
When asked to evaluate complex vector expressions containing a · ( × )$\vec{a} \cdot (\dots \times \dots)$, immediately distribute and convert them into Scalar Triple Products [ a b c]$[\vec{a} \ \vec{b} \ \vec{c}]$. Cyclic permutations and given cross-product relationships will rapidly collapse the expression into simple magnitudes.
Chapter Mix
Class 12 Maths: Vector Algebra
