Vector Algebra
Finding a vector from dot and cross product conditions
nta_pyq_2023_jan
Grade 12
Question:
Let $\vec{a} = -\hat{i}-\hat{j}+\hat{k}$, $\vec{a}\cdot\vec{b} = 1$ and $\vec{a}\times\vec{b} = \hat{i}-\hat{j}$. Then $\vec{a}-6\vec{b}$ is equal to
3(\hat{i}-\hat{j}-\hat{k})
3(\hat{i}+\hat{j}+\hat{k})
3(\hat{i}-\hat{j}+\hat{k})
3(\hat{i}+\hat{j}-\hat{k})
Step-by-Step Solution
Key Concept: Take cross product of both sides of $\vec{a}\times\vec{b} = \hat{i}-\hat{j}$ with $\vec{a}$, use BAC-CAB and the given dot product to find $\vec{b}$.
$\vec{a}\times(\hat{i}-\hat{j}) = \hat{i}(1+1)+\hat{j}(1+1)+\hat{k}(1+1) = \hat{i}+\hat{j}+2\hat{k}$. Wait: $\vec{a} = (-1,-1,1)$, $\vec{a}\times(\hat{i}-\hat{j}) = \begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\\-1&-1&1\\1&-1&0\end{vmatrix} = (1)\hat{i}+\hat{j}+2\hat{k}$. So $\vec{a}-3\vec{b} = \hat{i}+\hat{j}+2\hat{k}$, $\vec{b} = (\vec{a}-\hat{i}-\hat{j}-2\hat{k})/3 = (-2\hat{i}-2\hat{j}-\hat{k})/3$. From solution: $\vec{a}-6\vec{b} = 3\hat{i}+3\hat{j}+3\hat{k} = 3(\hat{i}+\hat{j}+\hat{k})$. Answer: (2)
Correct Answer: $3(\hat{i}+\hat{j}+\hat{k})$