To find the projection vector of \(\vec{f}\) on \(\vec{g}\), we use the formula for the projection of vector \(\vec{a}\) onto vector \(\vec{b}\):
\(\text{proj}_{\vec{b}}\vec{a}=\left(\frac{\vec{a}\cdot\vec{b}}{\vec{b}\cdot\vec{b}}\right)\vec{b}\)
Given \(\vec{f}=i+j+k\) and \(\vec{g}=2i-j+3k\), we first calculate the dot product \(\vec{f}\cdot\vec{g}\):
\(\vec{f}\cdot\vec{g}=(i+j+k)\cdot(2i-j+3k)=1\cdot2+1\cdot(-1)+1\cdot3=2-1+3=4\)
Next, compute \(\vec{g}\cdot\vec{g}\):
\(\vec{g}\cdot\vec{g}=(2i-j+3k)\cdot(2i-j+3k)=2^2+(-1)^2+3^2=4+1+9=14\)
Now substitute these into the projection formula:
\(\text{proj}_{\vec{g}}\vec{f}=\left(\frac{4}{14}\right)(2i-j+3k)=\frac{2}{7}(2i-j+3k)\)
Thus, the projection vector of \(\vec{f}\) on \(\vec{g}\) is \(\frac{2}{7}(2i-j+3k)\).
We are given two vectors: \[ \vec{f} = \hat{i} + \hat{j} + \hat{k}, \quad \vec{g} = 2\hat{i} - \hat{j} + 3\hat{k} \] Step 1: Recall the Formula for Projection The projection vector of \( \vec{f} \) on \( \vec{g} \) is given by: \[ \text{Proj}_{\vec{g}} \vec{f} = \frac{\vec{f} \cdot \vec{g}}{|\vec{g}|^2} \vec{g} \] Step 2: Compute the Dot Product \[ \vec{f} \cdot \vec{g} = (1)(2) + (1)(-1) + (1)(3) \] \[ \vec{f} \cdot \vec{g} = 2 - 1 + 3 = 4 \] Step 3: Compute \( |\vec{g}|^2 \) \[ |\vec{g}|^2 = (2)^2 + (-1)^2 + (3)^2 \] \[ |\vec{g}|^2 = 4 + 1 + 9 = 14 \] Step 4: Compute the Projection Vector \[ \text{Proj}_{\vec{g}} \vec{f} = \frac{4}{14} \vec{g} = \frac{2}{7} \vec{g} \] \[ = \frac{2}{7} (2\hat{i} - \hat{j} + 3\hat{k}) \] Step 5: Final Answer
\[Correct Answer: (2) \ \frac{2}{7} (2\hat{i} - \hat{j} + 3\hat{k})\]Let \( \vec{p} \) and \( \vec{q} \) be two unit vectors and \( \alpha \) be the angle between them. Then \( (\vec{p} + \vec{q}) \) will be a unit vector for what value of \( \alpha \)?