Question

Two position vectors are given by \[ \vec{r}_1=(1,1,1) \] and \[ \vec{r}_2=(1,-1,1) \] The unit vector in the direction of \[ \vec{r}_1\times \vec{r}_2 \] is

• A. \(\dfrac{\hat{i}}{\sqrt{2}}-\dfrac{\hat{k}}{\sqrt{2}}\)
• B. \(-\dfrac{\hat{i}}{\sqrt{2}}-\dfrac{\hat{k}}{\sqrt{2}}\)
• C. \(\dfrac{\hat{i}}{\sqrt{2}}+\dfrac{\hat{k}}{\sqrt{2}}\)
• D. \(-\dfrac{\hat{i}}{\sqrt{2}}+\dfrac{\hat{k}}{\sqrt{2}}\)

Hint:

To find a unit vector in the direction of a vector \(\vec{a}\), use \[ \hat{a}=\frac{\vec{a}}{|\vec{a}|} \] Always calculate the cross product first and then divide by its magnitude.

Answer 1

The Correct Option is A

Step 1: Write the vectors in determinant form.
Given, \[ \vec{r}_1=\hat{i}+\hat{j}+\hat{k} \] and \[ \vec{r}_2=\hat{i}-\hat{j}+\hat{k} \] Now, \[ \vec{r}_1\times \vec{r}_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 1 & 1 \\ 1 & -1 & 1 \end{vmatrix} \]

Step 2: Expand the determinant.
Expanding along the first row, \[ \vec{r}_1\times \vec{r}_2 = \hat{i} \begin{vmatrix} 1 & 1 \\ -1 & 1 \end{vmatrix} - \hat{j} \begin{vmatrix} 1 & 1 \\ 1 & 1 \end{vmatrix} + \hat{k} \begin{vmatrix} 1 & 1 \\ 1 & -1 \end{vmatrix} \] Now calculate each determinant: \[ \begin{vmatrix} 1 & 1 \\ -1 & 1 \end{vmatrix} = 1(1)-1(-1) = 2 \] \[ \begin{vmatrix} 1 & 1 \\ 1 & 1 \end{vmatrix} = 1(1)-1(1) = 0 \] \[ \begin{vmatrix} 1 & 1 \\ 1 & -1 \end{vmatrix} = 1(-1)-1(1) = -2 \] Therefore, \[ \vec{r}_1\times \vec{r}_2 = 2\hat{i}-2\hat{k} \]

Step 3: Find the magnitude of the vector.
Magnitude: \[ \left|2\hat{i}-2\hat{k}\right| = \sqrt{2^2+(-2)^2} \] \[ = \sqrt{4+4} \] \[ = \sqrt{8} = 2\sqrt{2} \]

Step 4: Find the unit vector.
The unit vector in the direction of \[ \vec{r}_1\times \vec{r}_2 \] is \[ \frac{2\hat{i}-2\hat{k}}{2\sqrt{2}} \] \[ = \frac{\hat{i}}{\sqrt{2}}-\frac{\hat{k}}{\sqrt{2}} \]

Step 5: Final conclusion.
Hence, the required unit vector is \[ \boxed{\frac{\hat{i}}{\sqrt{2}}-\frac{\hat{k}}{\sqrt{2}}} \]