Core Logic
Evaluate base vectors:
a + b = (5, 1, 4)$$\vec{a} + \vec{b} = (5, 1, 4)$$
a × b = vmatrix i & j & k 3 & 2 & 1 2 & -1 & 3 vmatrix = (7, -7, -7)$$\vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & 2 & 1 \\ 2 & -1 & 3 \end{vmatrix} = (7, -7, -7)$$
Substitute into given cross product equation, letting c = (x, y, z)$\vec{c} = (x, y, z)$:
(5 i + j + 4 k) × (x i + y j + z k) = 2(7 i - 7 j - 7 k) + 24 j - 6 k$$(5\hat{i} + \hat{j} + 4\hat{k}) \times (x\hat{i} + y\hat{j} + z\hat{k}) = 2(7\hat{i} - 7\hat{j} - 7\hat{k}) + 24\hat{j} - 6\hat{k}$$
vmatrix i & j & k 5 & 1 & 4 x & y & z vmatrix = (14, -14, -14) + (0, 24, -6)$$\begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 5 & 1 & 4 \\ x & y & z \end{vmatrix} = (14, -14, -14) + (0, 24, -6)$$
(z-4y) i - (5z-4x) j + (5y-x) k = (14, 10, -20)$$(z-4y)\hat{i} - (5z-4x)\hat{j} + (5y-x)\hat{k} = (14, 10, -20)$$
Equating components:
z - 4y = 14 z = 4y + 14$$z - 4y = 14 \implies z = 4y + 14$$
4x - 5z = 10$4x - 5z = 10$
5y - x = -20 x = 5y + 20$$5y - x = -20 \implies x = 5y + 20$$
Now use the dot product constraint:
a - b + i = (3-2+1) i + (2+1) j + (1-3) k = (2, 3, -2)$\vec{a} - \vec{b} + \hat{i} = (3-2+1)\hat{i} + (2+1)\hat{j} + (1-3)\hat{k} = (2, 3, -2)$.
( a - b + i) · c = -3$$(\vec{a} - \vec{b} + \hat{i}) \cdot \vec{c} = -3$$
2x + 3y - 2z = -3$$2x + 3y - 2z = -3$$
Substitute x = 5y + 20$x = 5y + 20$ and z = 4y + 14$z = 4y + 14$:
2(5y + 20) + 3y - 2(4y + 14) = -3$$2(5y + 20) + 3y - 2(4y + 14) = -3$$
10y + 40 + 3y - 8y - 28 = -3$$10y + 40 + 3y - 8y - 28 = -3$$
5y + 12 = -3 5y = -15 y = -3$$5y + 12 = -3 \implies 5y = -15 \implies y = -3$$
Calculate x$x$ and z$z$:
x = 5(-3) + 20 = 5$$x = 5(-3) + 20 = 5$$
z = 4(-3) + 14 = 2$$z = 4(-3) + 14 = 2$$
So, c = (5, -3, 2)$\vec{c} = (5, -3, 2)$.
| c|² = 5² + (-3)² + 2² = 25 + 9 + 4 = 38$$|\vec{c}|^2 = 5^2 + (-3)^2 + 2^2 = 25 + 9 + 4 = 38$$
Chapter Mix
Class 12 Maths: Vector Algebra
