Question

A plane passes through the points A(1, 2, 3), B(2, 3, 1) and C(2, 4, 2). If O is the origin and P is (2, -1, 1), then the projection of $\vec{OP}$ on this plane is of length :

• A. $\sqrt{\frac{2}{3}}$
• B. $\sqrt{\frac{2}{11}}$
• C. $\sqrt{\frac{2}{7}}$
• D. $\sqrt{\frac{2}{5}}$

Hint:

The length of the projection of a vector $\vec{v}$ onto a plane with normal $\vec{n}$ can be found using a form of the Pythagorean theorem in 3D: $|\vec{v}|^2 = (\text{length of projection on plane})^2 + (\text{length of projection on normal})^2$. It's often easier to calculate the projection on the normal first.

Answer 1

The Correct Option is B

The plane passes through points $A(1,2,3)$, $B(2,3,1)$ and $C(2,4,2)$. 
Step 1: Find direction vectors in the plane \[ \vec{AB} = (2-1,\,3-2,\,1-3) = (1,1,-2) \] \[ \vec{AC} = (2-1,\,4-2,\,2-3) = (1,2,-1) \] Step 2: Find the normal vector to the plane 

Step 3: Vector to be projected \[ \vec{OP} = (2,-1,1) \] Step 4: Use projection formula \[ |\vec{OP}|^2 = 2^2 + (-1)^2 + 1^2 = 6 \] \[ \vec{OP} \cdot \vec{n} = (2)(3)+(-1)(-1)+(1)(1)=8 \] \[ |\vec{n}|^2 = 3^2+(-1)^2+1^2 = 11 \] \[ \text{Projection length}^2 = |\vec{OP}|^2 - \left(\frac{\vec{OP}\cdot\vec{n}}{|\vec{n}|}\right)^2 = 6 - \frac{64}{11} = \frac{2}{11} \] \[ \boxed{\text{Projection length}=\sqrt{\frac{2}{11}}} \]