Question

What will be the projection of Vector IJK on IJ

Answer 1

I+J

Answer 2

To find the projection of vector $\vec{A}$ onto vector $\vec{B}$, we use the formula:

$$\text{proj}_{\vec{B}} \vec{A} = \frac{\vec{A} \cdot \vec{B}}{|\vec{B}|^2} \vec{B}$$

First, we interpret the given vectors:

Vector IJK is interpreted as $\vec{A} = \vec{I} + \vec{J} + \vec{K}$.

Vector IJ is interpreted as $\vec{B} = \vec{I} + \vec{J}$.

Now, we proceed with the calculation:

1. Calculate the dot product $\vec{A} \cdot \vec{B}$:

   $$\vec{A} \cdot \vec{B} = (\vec{I} + \vec{J} + \vec{K}) \cdot (\vec{I} + \vec{J})$$

   Using the properties of unit vectors ($\vec{I} \cdot \vec{I} = 1$, $\vec{J} \cdot \vec{J} = 1$, $\vec{K} \cdot \vec{K} = 1$, and $\vec{I} \cdot \vec{J} = 0$, etc.):

   $$\vec{A} \cdot \vec{B} = (1)(1) + (1)(1) + (1)(0) = 1 + 1 + 0 = 2$$

2. Calculate the square of the magnitude of $\vec{B}$, i.e., $|\vec{B}|^2$:

   $$\vec{B} = \vec{I} + \vec{J}$$ $$|\vec{B}| = \sqrt{1^2 + 1^2} = \sqrt{1 + 1} = \sqrt{2}$$ $$|\vec{B}|^2 = (\sqrt{2})^2 = 2$$

3. Substitute these values into the projection formula:

   $$\text{proj}_{\vec{B}} \vec{A} = \frac{\vec{A} \cdot \vec{B}}{|\vec{B}|^2} \vec{B}$$ $$\text{proj}_{\vec{B}} \vec{A} = \frac{2}{2} (\vec{I} + \vec{J})$$ $$\text{proj}_{\vec{B}} \vec{A} = 1 \cdot (\vec{I} + \vec{J})$$ $$\text{proj}_{\vec{B}} \vec{A} = \vec{I} + \vec{J}$$

Geometrically, projecting $\vec{A} = \vec{I} + \vec{J} + \vec{K}$ (vector to (1,1,1)) onto $\vec{B} = \vec{I} + \vec{J}$ (vector to (1,1,0) in the XY-plane) means finding the foot of the perpendicular from the tip of $\vec{A}$ to the line containing $\vec{B}$. Since $\vec{B}$ lies in the XY-plane and $\vec{A}$ has a Z-component of 1, dropping a perpendicular from (1,1,1) onto the XY-plane results in the point (1,1,0), which is precisely the tip of $\vec{B}$.