Question

The distance between a point \(P\) whose position vector is \[ 5\vec{i}+\vec{j}+3\vec{k} \] and the line \[ \vec{r}=(3\vec{i}+7\vec{j}+\vec{k})+t(\vec{j}+\vec{k}) \] is:

• A. \(3\)
• B. \(4\)
• C. \(5\)
• D. \(6\)

Hint:

For the distance between a point and a line in 3D, \[ d=\frac{|(\overrightarrow{AP}\times \vec{d})|}{|\vec{d}|}, \] where \(\vec{d}\) is the direction vector of the line and \(A\) is any point on the line.

Answer 1

The Correct Option is D

Step 1: Identify a point on the line and its direction vector.
The line is \[ \vec{r}=(3\vec{i}+7\vec{j}+\vec{k})+t(\vec{j}+\vec{k}). \] Hence, a point on the line is \[ A(3,7,1) \] and the direction vector is \[ \vec{d}=\hat{j}+\hat{k}=(0,1,1). \] The given point is \[ P(5,1,3). \]

Step 2: Form the vector \(\overrightarrow{AP}\).
\[ \overrightarrow{AP} = (5-3)\hat{i}+(1-7)\hat{j}+(3-1)\hat{k}. \] \[ \overrightarrow{AP} = 2\hat{i}-6\hat{j}+2\hat{k}. \]

Step 3: Use the distance formula from a point to a line.
The distance of point \(P\) from the line is \[ d= \frac{\left|\overrightarrow{AP}\times \vec{d}\right|} {|\vec{d}|}. \] Compute the cross product: \[ \overrightarrow{AP}\times \vec{d} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} 2 & -6 & 2 0 & 1 & 1 \end{vmatrix}. \] \[ = \hat{i}(-6-2) -\hat{j}(2-0) +\hat{k}(2-0). \] \[ = -8\hat{i}-2\hat{j}+2\hat{k}. \]

Step 4: Find its magnitude.
\[ \left|\overrightarrow{AP}\times \vec{d}\right| = \sqrt{(-8)^2+(-2)^2+2^2}. \] \[ = \sqrt{64+4+4}. \] \[ = \sqrt{72} = 6\sqrt{2}. \] Also, \[ |\vec{d}| = \sqrt{0^2+1^2+1^2} = \sqrt{2}. \]

Step 5: Calculate the distance.
\[ d = \frac{6\sqrt{2}}{\sqrt{2}} = 6. \]

Step 6: Final conclusion.
Therefore, \[ \boxed{6} \]