Let veca = -hati + 2hatj + 2hatk$\vec{a} = -\hat{i} + 2\hat{j} + 2\hat{k}$ , vecb = 8hati + 7hatj - 3hatk$\vec{b} = 8\hat{i} + 7\hat{j} - 3\hat{k}$ and vecc$\vec{c}$ be a vector such that veca times vecc = vecb$\vec{a} \times \vec{c} = \vec{b}$ . If vecc. (hati + hatj + hatk) = 4$\vec{c}. (\hat{i} + \hat{j} + \hat{k}) = 4$ , then |veca + vecc|^2$|\vec{a} + \vec{c}|^2$ is equal to:

### Related Formula veca times vecc = beginvmatrix hati & hatj & hatk \\ a_1 & a_2 & a_3 \\ c_1 & c_2 & c_3 endvmatrix$$\vec{a} \times \vec{c} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ a_1 & a_2 & a_3 \\ c_1 & c_2 & c_3 \end{vmatrix}$$ ### Core Logic Let the unknown vector be vecc = c_1hati + c_2hatj + c_3hatk$\vec{c} = c_1\hat{i} + c_2\hat{j} + c_3\hat{k}$. Given veca = -hati + 2hatj + 2hatk$\vec{a} = -\hat{i} + 2\hat{j} + 2\hat{k}$ and vecb = 8hati + 7hatj - 3hatk$\vec{b} = 8\hat{i} + 7\hat{j} - 3\hat{k}$. Compute the cross product veca times vecc$\vec{a} \times \vec{c}$: veca times vecc = (2c_3 - 2c_2)hati + (-1 cdot c_3 - 2c_1)(-1)hatj + (-1 cdot c_2 - 2c_1)hatk$$\vec{a} \times \vec{c} = (2c_3 - 2c_2)\hat{i} + (-1 \cdot c_3 - 2c_1)(-1)\hat{j} + (-1 \cdot c_2 - 2c_1)\hat{k}$$ = (2c_3 - 2c_2)hati + (c_3 + 2c_1)hatj - (c_2 + 2c_1)hatk$$= (2c_3 - 2c_2)\hat{i} + (c_3 + 2c_1)\hat{j} - (c_2 + 2c_1)\hat{k}$$ ### Step 1: Equate components to find c Equating this to vecb$\vec{b}$: 2c_3 - 2c_2 = 8 Rightarrow c_3 - c_2 = 4$2c_3 - 2c_2 = 8 \Rightarrow c_3 - c_2 = 4$ c_3 + 2c_1 = 7$c_3 + 2c_1 = 7$ -(c_2 + 2c_1) = -3 Rightarrow c_2 + 2c_1 = 3$-(c_2 + 2c_1) = -3 \Rightarrow c_2 + 2c_1 = 3$ We are also given vecc cdot (hati + hatj + hatk) = 4$\vec{c} \cdot (\hat{i} + \hat{j} + \hat{k}) = 4$: c_1 + c_2 + c_3 = 4$$c_1 + c_2 + c_3 = 4$$

### Step 2: Solve the linear system From the dot product equation: c_1 = 4 - c_2 - c_3$c_1 = 4 - c_2 - c_3$. Substitute into c_2 + 2c_1 = 3$c_2 + 2c_1 = 3$: c_2 + 2(4 - c_2 - c_3) = 3 Rightarrow 8 - c_2 - 2c_3 = 3 Rightarrow c_2 + 2c_3 = 5$$c_2 + 2(4 - c_2 - c_3) = 3 \Rightarrow 8 - c_2 - 2c_3 = 3 \Rightarrow c_2 + 2c_3 = 5$$ We have the system: c_3 - c_2 = 4 Rightarrow c_2 = c_3 - 4$c_3 - c_2 = 4 \Rightarrow c_2 = c_3 - 4$ Substitute into above: (c_3 - 4) + 2c_3 = 5 Rightarrow 3c_3 = 9 Rightarrow c_3 = 3$(c_3 - 4) + 2c_3 = 5 \Rightarrow 3c_3 = 9 \Rightarrow c_3 = 3$ Then c_2 = 3 - 4 = -1$c_2 = 3 - 4 = -1$. Then c_1 = 4 - (-1) - 3 = 2$c_1 = 4 - (-1) - 3 = 2$. So, vecc = 2hati - hatj + 3hatk$\vec{c} = 2\hat{i} - \hat{j} + 3\hat{k}$. ### Step 3: Evaluate the final expression veca + vecc = (-hati + 2hatj + 2hatk) + (2hati - hatj + 3hatk) = hati + hatj + 5hatk$$\vec{a} + \vec{c} = (-\hat{i} + 2\hat{j} + 2\hat{k}) + (2\hat{i} - \hat{j} + 3\hat{k}) = \hat{i} + \hat{j} + 5\hat{k}$$ |veca + vecc|^2 = 1^2 + 1^2 + 5^2 = 1 + 1 + 25 = 27$$|\vec{a} + \vec{c}|^2 = 1^2 + 1^2 + 5^2 = 1 + 1 + 25 = 27$$ ### Pattern Recognition Whenever veca times vecc = vecb$\vec{a} \times \vec{c} = \vec{b}$ and a dot product constraint is given, simply expand vecc$\vec{c}$ algebraically as (c_1, c_2, c_3)$(c_1, c_2, c_3)$. The cross product and dot product create an easily solvable 3x3 linear system. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Maths: Vector Algebra