Question:
A vector \(\vec{a}\) of magnitude 14 has direction ratios \(\langle 2, 3, -6 \rangle\). Find the projection of the vector \(\vec{a}\) on \(\hat{i}\).
A vector \(\vec{a}\) of magnitude 14 has direction ratios \(\langle 2, 3, -6 \rangle\). Find the projection of the vector \(\vec{a}\) on \(\hat{i}\).
Show Hint
Divide the direction ratios by 7 to get direction cosines, then projection on i-hat equals 14 times the first direction cosine.
Updated On: Oct 22, 2026
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Solution and Explanation
Step 1: Understanding the Concept:
The projection of \(\vec{a}\) on \(\hat{i}\) is \(\dfrac{\vec{a} \cdot \hat{i}}{|\hat{i}|}\). Since \(|\hat{i}| = 1\), it equals \(\vec{a} \cdot \hat{i}\), the x-component of \(\vec{a}\).
Step 2: Convert the direction ratios into direction cosines.
The direction ratios are \(2, 3, -6\).
\[ \sqrt{2^2 + 3^2 + (-6)^2} = \sqrt{4 + 9 + 36} = \sqrt{49} = 7 \]
So the direction cosines are \(\dfrac{2}{7}, \dfrac{3}{7}, \dfrac{-6}{7}\).
Step 3: Write the vector.
The vector is its magnitude times the unit vector in its direction: \[ \vec{a} = 14\left(\frac{2}{7}\hat{i} + \frac{3}{7}\hat{j} - \frac{6}{7}\hat{k}\right) = 4\hat{i} + 6\hat{j} - 12\hat{k} \]
Step 4: Find the projection.
\[ \vec{a} \cdot \hat{i} = (4)(1) + (6)(0) + (-12)(0) = 4 \]
Final Answer:
The projection of \(\vec{a}\) on \(\hat{i}\) is 4 units. \[ \boxed{4} \]
The projection of \(\vec{a}\) on \(\hat{i}\) is \(\dfrac{\vec{a} \cdot \hat{i}}{|\hat{i}|}\). Since \(|\hat{i}| = 1\), it equals \(\vec{a} \cdot \hat{i}\), the x-component of \(\vec{a}\).
Step 2: Convert the direction ratios into direction cosines.
The direction ratios are \(2, 3, -6\).
\[ \sqrt{2^2 + 3^2 + (-6)^2} = \sqrt{4 + 9 + 36} = \sqrt{49} = 7 \]
So the direction cosines are \(\dfrac{2}{7}, \dfrac{3}{7}, \dfrac{-6}{7}\).
Step 3: Write the vector.
The vector is its magnitude times the unit vector in its direction: \[ \vec{a} = 14\left(\frac{2}{7}\hat{i} + \frac{3}{7}\hat{j} - \frac{6}{7}\hat{k}\right) = 4\hat{i} + 6\hat{j} - 12\hat{k} \]
Step 4: Find the projection.
\[ \vec{a} \cdot \hat{i} = (4)(1) + (6)(0) + (-12)(0) = 4 \]
Final Answer:
The projection of \(\vec{a}\) on \(\hat{i}\) is 4 units. \[ \boxed{4} \]
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