Let veca=-5hati+hatj-3hatk$\vec{a}=-5\hat{i}+\hat{j}-3\hat{k}$, vecb=hati+2hatj-4hatk$\vec{b}=\hat{i}+2\hat{j}-4\hat{k}$ and vecc=(((vecatimesvecb)timeshati)timeshati)timeshati.$\vec{c}=(((\vec{a}\times\vec{b})\times\hat{i})\times\hat{i})\times\hat{i}.$ Then vecccdot(-hati+hatj+hatk)$\vec{c}\cdot(-\hat{i}+\hat{j}+\hat{k})$ is equal to:

### Related Formula Vector Triple Product Identity: (vecu times vecv) times vecw = (vecu cdot vecw)vecv - (vecv cdot vecw)veca$$(\vec{u} \times \vec{v}) \times \vec{w} = (\vec{u} \cdot \vec{w})\vec{v} - (\vec{v} \cdot \vec{w})\vec{a}$$ ### Core Logic Given vectors veca$\vec{a}$ and vecb$\vec{b}$: First, let's find the inner vector triple product (veca times vecb) times hati$(\vec{a} \times \vec{b}) \times \hat{i}$: (veca times vecb) times hati = (veca cdot hati)vecb - (vecb cdot hati)veca $$(\vec{a} \times \vec{b}) \times \hat{i} = (\vec{a} \cdot \hat{i})\vec{b} - (\vec{b} \cdot \hat{i})\vec{a} $$ From the given components: - veca cdot hati = -5$\vec{a} \cdot \hat{i} = -5$ - vecb cdot hati = 1$\vec{b} \cdot \hat{i} = 1$ (veca times vecb) times hati = -5vecb - veca $$(\vec{a} \times \vec{b}) \times \hat{i} = -5\vec{b} - \vec{a} $$ ### Step 1: Substitute Vector Coordinates Substitute coordinates of veca$\vec{a}$ and vecb$\vec{b}$ into the expression: -5vecb - veca = -5(hati + 2hatj - 4hatk) - (-5hati + hatj - 3hatk)$$-5\vec{b} - \vec{a} = -5(\hat{i} + 2\hat{j} - 4\hat{k}) - (-5\hat{i} + \hat{j} - 3\hat{k})$$ = -5hati - 10hatj + 20hatk + 5hati - hatj + 3hatk = -11hatj + 23hatk $$= -5\hat{i} - 10\hat{j} + 20\hat{k} + 5\hat{i} - \hat{j} + 3\hat{k} = -11\hat{j} + 23\hat{k} $$ Now, perform successive cross products with hati$\hat{i}$: textNext Step = (-11hatj + 23hatk) times hati = -11(hatj times hati) + 23(hatk times hati) = 11hatk + 23hatj $$\text{Next Step} = (-11\hat{j} + 23\hat{k}) \times \hat{i} = -11(\hat{j} \times \hat{i}) + 23(\hat{k} \times \hat{i}) = 11\hat{k} + 23\hat{j} $$ textFinal Vector vecc = (23hatj + 11hatk) times hati = 23(hatj times hati) + 11(hatk times hati) = -23hatk + 11hatj $$\text{Final Vector } \vec{c} = (23\hat{j} + 11\hat{k}) \times \hat{i} = 23(\hat{j} \times \hat{i}) + 11(\hat{k} \times \hat{i}) = -23\hat{k} + 11\hat{j} $$ ### Step 2: Calculate Dot Product Now evaluate vecc cdot (-hati + hatj + hatk)$\vec{c} \cdot (-\hat{i} + \hat{j} + \hat{k})$: vecc = 11hatj - 23hatk $$\vec{c} = 11\hat{j} - 23\hat{k} $$ vecc cdot (-hati + hatj + hatk) = (0)( -1) + (11)(1) + (-23)(1) = 11 - 23 = -12 $$\vec{c} \cdot (-\hat{i} + \hat{j} + \hat{k}) = (0)( -1) + (11)(1) + (-23)(1) = 11 - 23 = -12 $$ ### Pattern Recognition Sees: Repeated vector cross products with standard unit vectors. Shortcut: Observe that crossing a vector in the y-z$y-z$ plane with hati$\hat{i}$ rotates its components by 90^circ$90^{\circ}$ in that plane. Doing it twice returns it to the original plane with flipped coefficients and signs (vecv times hati times hati = -vecv$\vec{v} \times \hat{i} \times \hat{i} = -\vec{v}$). ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Mathematics: Vector Algebra