If {a} is a nonzero vector such that its projections on the vectors 2{i}-{j}+2{k}, {i}+2{j}-2{k} and {k} are equal, then a unit vector along {a} is [2 Apr 2025 Shift 1]

Correct Answer is : 1155(7i^+9j^+5k^){1}{{155}}(7{i}+9{j}+5{k})1551(7i^+9j^+5k^)

Correct Answer is : 1155(7i^+9j^+5k^){1}{{155}}(7{i}+9{j}+5{k})1551(7i^+9j^+5k^)

Test Index

Vector Algebra Part 3

Section: Mathematics

Question : 27 of 72

Marks: +1, -0

\vec{a}

2\hat{i}-\hat{j}+2\hat{k}, \hat{i}+2\hat{j}-2\hat{k}

\hat{k}

\vec{a}

Solution:

Projection of

\vec{a}

\vec{v}

= \frac{\vec{a} \cdot \vec{v}}{|\vec{v}|} -

\Rightarrow \frac{\vec{a} \cdot (2\hat{i}-\hat{j}+2\hat{k})}{3} = \frac{\vec{a} \cdot \hat{k}}{1} = \frac{\vec{a} \cdot (\hat{i}+2\hat{j}-2\hat{k})}{3}

\Rightarrow \vec{a} \cdot (2\hat{i}-\hat{j}-\hat{k})=0 \text{ and } \vec{a} \cdot (\hat{i}+2\hat{j}-5\hat{k})=0

\Rightarrow \vec{a} \perp (2\hat{i}-\hat{j}-\hat{k}) \text{ and } (\hat{i}+2\hat{j}-5\hat{k})

\Rightarrow \vec{a} \parallel (2\hat{i}-\hat{j}-\hat{k}) \times (\hat{i}+2\hat{j}-5\hat{k})

\Rightarrow \vec{a} = \pm k \begin{vmatrix} \hat{i} & -\hat{j} & \hat{k} \\ 2 & -1 & -1 \\ 1 & 2 & -5 \end{vmatrix} = \pm k (7\hat{i}+9\hat{j}-5\hat{k})

⇒ Unit vector will be

\frac{1}{\sqrt{155}}(7\hat{i}+9\hat{j}+5\hat{k})

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