Question:
A straight line passes through the points \( (10,8,6) \) and \( (13,9,4) \). A unit vector parallel to this line is
A straight line passes through the points \( (10,8,6) \) and \( (13,9,4) \). A unit vector parallel to this line is
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Unit vector = direction vector divided by its magnitude.
Updated On: Apr 21, 2026
- \( \frac{1}{\sqrt{17}}(3\hat{i} + 2\hat{j} + 2\hat{k}) \)
- \( \frac{1}{\sqrt{6}}(\hat{i} + \hat{j} - 2\hat{k}) \)
- \( \frac{1}{\sqrt{14}}(3\hat{i} + \hat{j} + 2\hat{k}) \)
- \( \frac{1}{\sqrt{11}}(3\hat{i} + \hat{j} + 2\hat{k}) \)
- \( \frac{1}{\sqrt{14}}(3\hat{i} + \hat{j} - 2\hat{k}) \)
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The Correct Option is
Solution and Explanation
Concept:
Direction vector = difference of points.
Step 1: Find direction vector.
\[ \vec{d} = (13-10, 9-8, 4-6) = (3,1,-2) \]
Step 2: Find magnitude.
\[ |\vec{d}| = \sqrt{9 + 1 + 4} = \sqrt{14} \]
Step 3: Unit vector.
\[ \frac{1}{\sqrt{14}}(3\hat{i} + \hat{j} - 2\hat{k}) \]
Step 1: Find direction vector.
\[ \vec{d} = (13-10, 9-8, 4-6) = (3,1,-2) \]
Step 2: Find magnitude.
\[ |\vec{d}| = \sqrt{9 + 1 + 4} = \sqrt{14} \]
Step 3: Unit vector.
\[ \frac{1}{\sqrt{14}}(3\hat{i} + \hat{j} - 2\hat{k}) \]
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