Question 43

Mathematics Scalar and Vector Products Easy

The length of the projection of \(\vec{a} = 2\hat{i} + 3\hat{j} + \hat{k}\) on \(\vec{b} = -2\hat{i} + \hat{j} + 2\hat{k}\), is equal to:

(A) \(\frac{1}{3}\)
(B) \(\frac{-1}{3}\)
(C) \(\frac{2}{3}\)
(D) \(\frac{-2}{3}\)
View Dynamic Solution & Explanation
Correct Solution: Option A

Step-by-step Solution:

Correct Answer: A ($\frac{1}{3}$)

The length of the projection of one vector onto another, also known as the scalar projection, is found by calculating the dot product of the two vectors and dividing by the magnitude of the vector being projected onto.


Step 1: The Formula for the Length of Projection

The length of the projection of a vector $\vec{a}$ onto a vector $\vec{b}$ is given by the formula:

$$ \text{Length} = \frac{\vec{a} \cdot \vec{b}}{|\vec{b}|} $$

We need to calculate the dot product ($\vec{a} \cdot \vec{b}$) and the magnitude of $\vec{b}$ ($|\vec{b}|$).


Step 2: Calculate the Dot Product ($\vec{a} \cdot \vec{b}$)

The given vectors are:

  • $\vec{a} = 2\hat{i} + 3\hat{j} + \hat{k}$
  • $\vec{b} = -2\hat{i} + \hat{j} + 2\hat{k}$

The dot product is calculated as follows:

$\vec{a} \cdot \vec{b} = (2)(-2) + (3)(1) + (1)(2)$

$\vec{a} \cdot \vec{b} = -4 + 3 + 2 = 1$


Step 3: Calculate the Magnitude of $\vec{b}$ ($|\vec{b}|$)

The magnitude of vector $\vec{b}$ is calculated as:

$|\vec{b}| = \sqrt{(-2)^2 + (1)^2 + (2)^2}$

$|\vec{b}| = \sqrt{4 + 1 + 4} = \sqrt{9} = 3$


Step 4: Calculate the Length of the Projection

Now we substitute the values from the previous steps into the formula:

$$ \text{Length} = \frac{\vec{a} \cdot \vec{b}}{|\vec{b}|} = \frac{1}{3} $$