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Q4:
Find the projection of the vector $\hat{i} + 3\hat{j} + 7\hat{k}$ on the vector $7\hat{i} - \hat{j} + 8\hat{k}$.

Solution :

Step 1: Identifying Given Vectors

Let the two given vectors be represented as $\vec{a}$ and $\vec{b}$:

  • $\vec{a} = 1\hat{i} + 3\hat{j} + 7\hat{k}$
  • $\vec{b} = 7\hat{i} - 1\hat{j} + 8\hat{k}$

Step 2: Defining the Projection Formula

The scalar projection of a vector $\vec{a}$ on a vector $\vec{b}$ is defined as the dot product of $\vec{a}$ and the unit vector in the direction of $\vec{b}$. [Per the definition of Scalar Projection]:

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

Step 3: Calculating the Dot Product ($\vec{a} \cdot \vec{b}$)**

The dot product of two vectors $\vec{a} = a_1\hat{i} + a_2\hat{j} + a_3\hat{k}$ and $\vec{b} = b_1\hat{i} + b_2\hat{j} + b_3\hat{k}$ is given by $a_1b_1 + a_2b_2 + a_3b_3$.

$\vec{a} \cdot \vec{b} = (1)(7) + (3)(-1) + (7)(8)$

$\vec{a} \cdot \vec{b} = 7 - 3 + 56$

$\vec{a} \cdot \vec{b} = 60$ [Arithmetic summation]

Step 4: Calculating the Magnitude of Vector $\vec{b}$ ($|\vec{b}|$)

The magnitude of a vector $\vec{b} = b_1\hat{i} + b_2\hat{j} + b_3\hat{k}$ is defined as $\sqrt{b_1^2 + b_2^2 + b_3^2}$.

$|\vec{b}| = \sqrt{(7)^2 + (-1)^2 + (8)^2}$

$|\vec{b}| = \sqrt{49 + 1 + 64}$

$|\vec{b}| = \sqrt{114}$

Step 5: Computing the Final Scalar Projection

Substituting the values derived in Steps 3 and 4 into the formula established in Step 2:

$\text{Projection} = \frac{60}{\sqrt{114}}$

To rationalize the denominator (optional, but standard practice):

$\frac{60}{\sqrt{114}} \times \frac{\sqrt{114}}{\sqrt{114}} = \frac{60\sqrt{114}}{114}$

Dividing numerator and denominator by 6:

$\text{Projection} = \frac{10\sqrt{114}}{19}$

a b Projection of a on b

Final Solution: $\frac{60}{\sqrt{114}}$ or $\frac{10\sqrt{114}}{19}$


More Questions from Class 12 Mathematics Vector Algebra EXERCISE 10.3


CBSE Solutions for Class 12 Mathematics Vector Algebra


Chapters in CBSE - Class 12 Mathematics


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