Related Formula
If X × Y = 0, then vectors are parallel: Y = λ X$$\text{If } \vec{X} \times \vec{Y} = \vec{0}, \text{ then vectors are parallel: } \vec{Y} = \lambda \vec{X}$$
Core Logic
From the given cross product relation:
2( a × c) + 3( b × c) = 0$$2(\vec{a} \times \vec{c}) + 3(\vec{b} \times \vec{c}) = \vec{0}$$
(2 a + 3 b) × c = 0$$(2\vec{a} + 3\vec{b}) \times \vec{c} = \vec{0}$$
This implies that c$\vec{c}$ is parallel to (2 a + 3 b)$(2\vec{a} + 3\vec{b})$.
c = λ (2 a + 3 b)$$\vec{c} = \lambda (2\vec{a} + 3\vec{b})$$
Step 1: Finding Vector c in terms of Lambda
Calculate (2 a + 3 b)$(2\vec{a} + 3\vec{b})$:
2 a = 4 i - 10 j + 10 k$$2\vec{a} = 4\hat{i} - 10\hat{j} + 10\hat{k}$$
3 b = 3 i - 3 j + 9 k$$3\vec{b} = 3\hat{i} - 3\hat{j} + 9\hat{k}$$
2 a + 3 b = 7 i - 13 j + 19 k$$2\vec{a} + 3\vec{b} = 7\hat{i} - 13\hat{j} + 19\hat{k}$$
So, c = λ(7 i - 13 j + 19 k)$\vec{c} = \lambda(7\hat{i} - 13\hat{j} + 19\hat{k})$.
Step 2: Solving for Lambda
Use the dot product condition: ( a - b) · c = -97$(\vec{a} - \vec{b}) \cdot \vec{c} = -97$
First, find a - b$\vec{a} - \vec{b}$:
a - b = (2 - 1) i + (-5 - (-1)) j + (5 - 3) k = i - 4 j + 2 k$$\vec{a} - \vec{b} = (2 - 1)\hat{i} + (-5 - (-1))\hat{j} + (5 - 3)\hat{k} = \hat{i} - 4\hat{j} + 2\hat{k}$$
Now, substitute c$\vec{c}$ into the dot product:
( i - 4 j + 2 k) · λ(7 i - 13 j + 19 k) = -97$$(\hat{i} - 4\hat{j} + 2\hat{k}) \cdot \lambda(7\hat{i} - 13\hat{j} + 19\hat{k}) = -97$$
λ(1(7) + (-4)(-13) + 2(19)) = -97$$\lambda(1(7) + (-4)(-13) + 2(19)) = -97$$
λ(7 + 52 + 38) = -97$$\lambda(7 + 52 + 38) = -97$$
97λ = -97 λ = -1$$97\lambda = -97 \implies \lambda = -1$$
Step 3: Finding Final Magnitude
Since λ = -1$\lambda = -1$, vector c$\vec{c}$ is:
c = -7 i + 13 j - 19 k$$\vec{c} = -7\hat{i} + 13\hat{j} - 19\hat{k}$$
Now evaluate c × k$\vec{c} \times \hat{k}$:
c × k = (-7 i + 13 j - 19 k) × k$$\vec{c} \times \hat{k} = (-7\hat{i} + 13\hat{j} - 19\hat{k}) \times \hat{k}$$
= -7( i × k) + 13( j × k) - 19( k × k)$$= -7(\hat{i} \times \hat{k}) + 13(\hat{j} \times \hat{k}) - 19(\hat{k} \times \hat{k})$$
= -7(- j) + 13( i) - 0 = 13 i + 7 j$$= -7(-\hat{j}) + 13(\hat{i}) - 0 = 13\hat{i} + 7\hat{j}$$
The square of its magnitude is:
| c × k|² = 13² + 7² = 169 + 49 = 218$$|\vec{c} \times \hat{k}|^2 = 13^2 + 7^2 = 169 + 49 = 218$$
Pattern Recognition
An expression like p( A × C) + q( B × C) = 0$p(\vec{A} \times \vec{C}) + q(\vec{B} \times \vec{C}) = \vec{0}$ instantly collapses to (p A + q B) × C = 0$(p\vec{A} + q\vec{B}) \times \vec{C} = \vec{0}$, proving C$\vec{C}$ is collinear with the combined vector.
Chapter Mix
Class 12 Maths: Vector Algebra
