Question:
The projection of the vector \(u=(2,-1,3) \in \mathbb{R}^3\) onto the vector \(v=(1,2,-1)\) of the vector space \(\mathbb{R}^3\) is ____.
The projection of the vector \(u=(2,-1,3) \in \mathbb{R}^3\) onto the vector \(v=(1,2,-1)\) of the vector space \(\mathbb{R}^3\) is ____.
Show Hint
Use the formula \(\text{proj}_v u = \frac{u\cdot v}{v\cdot v}v\).
Updated On: Jul 3, 2026
- \((-1/2,-1,1/2)\)
- \((1/3,2/3,-1/3)\)
- \((2/3,4/3,-2/3)\)
- \((1/5,2/5,-1/5)\)
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The Correct Option is A
Solution and Explanation
Step 1: The projection of vector \(u\) onto vector \(v\) is given by \(\text{proj}_v u = \dfrac{u\cdot v}{v\cdot v}\,v\).
Step 2: Compute the dot product \(u \cdot v = (2)(1) + (-1)(2) + (3)(-1) = 2 - 2 - 3 = -3\).
Step 3: Compute \(v \cdot v = 1^2 + 2^2 + (-1)^2 = 1 + 4 + 1 = 6\).
Step 4: So the scalar factor is \(\dfrac{u\cdot v}{v \cdot v} = \dfrac{-3}{6} = -\dfrac{1}{2}\).
Step 5: Therefore \(\text{proj}_v u = -\dfrac{1}{2}(1,2,-1) = \left(-\dfrac{1}{2}, -1, \dfrac{1}{2}\right)\).
\[\boxed{\left(-\frac{1}{2},\,-1,\,\frac{1}{2}\right)}\]Was this answer helpful?
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