Question

Let \(\vec{a} = \hat{i} + 2\hat{j} + 3\hat{k}\), \(\vec{b} = 2\hat{i} + 3\hat{j} - 5\hat{k}\), and \(\vec{c} = 3\hat{i} - \hat{j} + \lambda\hat{k}\) be three vectors. Let \(\vec{r}\) be a unit vector along \(\vec{b} + \vec{c}\). If \(\vec{r} \cdot \vec{a} = 3\), then \(3\lambda\) is equal to:

• A. 27
• B. 25
• C. 25
• D. 21

Answer 1

The Correct Option is B

To ascertain the value of \(3\lambda\), we are provided with vectors \(\vec{a}\), \(\vec{b}\), and \(\vec{c}\), along with the condition \(\vec{r} \cdot \vec{a} = 3\). Here, \(\vec{r}\) signifies a unit vector collinear with \(\vec{b} + \vec{c}\).

1. The initial step involves calculating the vector sum \(\vec{b} + \vec{c}\):

\(\vec{b} = 2\hat{i} + 3\hat{j} - 5\hat{k}\) and \(\vec{c} = 3\hat{i} - \hat{j} + \lambda\hat{k}\).

Consequently, \[ \vec{b} + \vec{c} = (2 + 3)\hat{i} + (3 - 1)\hat{j} + (-5 + \lambda)\hat{k} = 5\hat{i} + 2\hat{j} + (\lambda - 5)\hat{k} \]

1. Next, we determine the magnitude of \(\vec{b} + \vec{c}\):

The magnitude is computed as:

1. \(|\vec{b} + \vec{c}| = \sqrt{5^2 + 2^2 + (\lambda - 5)^2}\)

\[ = \sqrt{25 + 4 + (\lambda^2 - 10\lambda + 25)} = \sqrt{\lambda^2 - 10\lambda + 54} \]

1. The unit vector \(\vec{r}\) in the direction of \(\vec{b} + \vec{c}\) is expressed as:

\[ \vec{r} = \frac{\vec{b} + \vec{c}}{|\vec{b} + \vec{c}|} = \frac{5\hat{i} + 2\hat{j} + (\lambda - 5)\hat{k}}{\sqrt{\lambda^2 - 10\lambda + 54}} \]

1. We are given the condition:

\(\vec{r} \cdot \vec{a} = 3\), where \(\vec{a} = \hat{i} + 2\hat{j} + 3\hat{k}\).

Substituting the expressions:

\[ \frac{(5\hat{i} + 2\hat{j} + (\lambda - 5)\hat{k}) \cdot (\hat{i} + 2\hat{j} + 3\hat{k})}{\sqrt{\lambda^2 - 10\lambda + 54}} = 3 \]

Performing the dot product simplification:

\[ (5)(1) + (2)(2) + ((\lambda - 5)(3)) = 5 + 4 + 3\lambda - 15 \]

\[ = 3\lambda - 6 \]

1. Equating the simplified dot product to 3 yields:

\[ \frac{3\lambda - 6}{\sqrt{\lambda^2 - 10\lambda + 54}} = 3 \]

To solve for \(\lambda\), we multiply both sides by \(\sqrt{\lambda^2 - 10\lambda + 54}\):

\[ 3\lambda - 6 = 3\sqrt{\lambda^2 - 10\lambda + 54} \]

Dividing the entire equation by 3 simplifies it to:

\[ \lambda - 2 = \sqrt{\lambda^2 - 10\lambda + 54} \]

1. Squaring both sides of the equation gives:

\[ (\lambda - 2)^2 = \lambda^2 - 10\lambda + 54 \]

Expanding the left side of the equation:

\[ \lambda^2 - 4\lambda + 4 = \lambda^2 - 10\lambda + 54 \]

After eliminating the \(\lambda^2\) terms and rearranging:

\[ -4\lambda + 4 = -10\lambda + 54 \]

\[ 6\lambda = 50 \]

This results in \(\lambda = \frac{50}{6} = \frac{25}{3}\).

1. Finally, we calculate the value of \(3\lambda\):

\[ 3\left(\frac{25}{3}\right) = 25 \]

Thus, the value of \(3\lambda\) is 25.