Question

Let\(\overrightarrow{ a }=2 \hat{i}-7 \hat{j}+5 \hat{k}, \overrightarrow{ b }=\hat{i}+\hat{k} and \overrightarrow{ c }=\hat{i}+2 \hat{j}-3 \hat{k}\) be three given vectors If \(\overrightarrow{ r }\) is a vector such that\( \vec{r} \times \vec{a}=\vec{c} \times \vec{a}  \ and \  \vec{r} \cdot \vec{b}=0,\) then \(|\vec{r}|\) is equal to :

• A. \(\frac{11}{7} \sqrt{2}\)
• B. \(\frac{11}{7}\)
• C. \(\frac{\sqrt{914}}{7}\)
• D. \(\frac{11}{5} \sqrt{2}\)

Answer 1

The Correct Option is A

The given vectors are:
\[\vec{a} = 2\hat{i} - 7\hat{j} + 5\hat{k}, \quad \vec{b} = \hat{i} + \hat{k}, \quad \vec{c} = \hat{i} + 2\hat{j} - 3\hat{k}.\]
Step 1: Use the Condition for \(\vec{r}\)
From \(\vec{r} \times \vec{a} = \vec{c} \times \vec{a}\), we can write:
\[(\vec{r} - \vec{c}) \times \vec{a} = 0.\]
This implies that \(\vec{r} - \vec{c}\) is parallel to \(\vec{a}\), so:
\[\vec{r} = \vec{c} + \lambda \vec{a},\]
where \(\lambda\) is a scalar.
Step 2: Use the Dot Product Condition
From \(\vec{r} \cdot \vec{b} = 0\), substitute \(\vec{r} = \vec{c} + \lambda \vec{a}\):
\[(\vec{c} + \lambda \vec{a}) \cdot \vec{b} = 0.\]
Expanding:
\[\vec{c} \cdot \vec{b} + \lambda (\vec{a} \cdot \vec{b}) = 0.\]
Calculate \(\vec{c} \cdot \vec{b}\):
\[\vec{c} \cdot \vec{b} = (\hat{i} + 2\hat{j} - 3\hat{k}) \cdot (\hat{i} + \hat{k}) = 1 + 0 - 3 = -2.\]
Calculate \(\vec{a} \cdot \vec{b}\):
\[\vec{a} \cdot \vec{b} = (2\hat{i} - 7\hat{j} + 5\hat{k}) \cdot (\hat{i} + \hat{k}) = 2 + 0 + 5 = 7.\]
Substitute into the equation:
\[-2 + \lambda (7) = 0.\]
Solve for \(\lambda\):
\[\lambda = \frac{2}{7}.\]
Step 3: Find \(\vec{r}\)
Substitute \(\lambda = \frac{2}{7}\) into \(\vec{r} = \vec{c} + \lambda \vec{a}\):
\[\vec{r} = (\hat{i} + 2\hat{j} - 3\hat{k}) + \frac{2}{7}(2\hat{i} - 7\hat{j} + 5\hat{k}).\]
Expand:
\[\vec{r} = \hat{i} + 2\hat{j} - 3\hat{k} + \frac{4}{7}\hat{i} - 2\hat{j} + \frac{10}{7}\hat{k}.\]
Combine terms:
\[\vec{r} = \left(1 + \frac{4}{7}\right)\hat{i} + \left(2 - 2\right)\hat{j} + \left(-3 + \frac{10}{7}\right)\hat{k}.\]
Simplify:
\[\vec{r} = \frac{11}{7}\hat{i} + 0\hat{j} + \frac{-11}{7}\hat{k}\].
Step 4: Find\(\lvert \vec{r} \rvert\)
The magnitude of \(\vec{r}\) is:
\[\lvert \vec{r} \rvert = \sqrt{\left(\frac{11}{7}\right)^2 + 0^2 + \left(\frac{-11}{7}\right)^2}.\]
Simplify:
\[\lvert \vec{r} \rvert = \sqrt{\frac{121}{49} + \frac{121}{49}} = \sqrt{\frac{242}{49}} = \frac{\sqrt{242}}{7}.\]
Simplify further:
\[\lvert \vec{r} \rvert = \frac{11}{7} \sqrt{2}.\]
Conclusion:
The magnitude of \(\vec{r}\) is \(\frac{11}{7} \sqrt{2}\).