Question

Consider two vectors $\vec{u} = 3\hat{i} - \hat{j}$ and $\vec{v} = 2\hat{i} + \hat{j} - \lambda \hat{k}$, $\lambda>0$. The angle between them is given by $\cos^{-1} \left( \frac{\sqrt{5}}{2\sqrt{7}} \right)$. Let $\vec{v} = \vec{v}_1 + \vec{v}_2$, where $\vec{v}_1$ is parallel to $\vec{u}$ and $\vec{v}_2$ is perpendicular to $\vec{u}$. Then the value $|\vec{v}_1|^2 + |\vec{v}_2|^2$ is equal to

• A. $\frac{23}{2}$
• B. 14
• C. $\frac{25}{2}$
• D. 10

Hint:

Use the dot product and magnitudes to find the angle between vectors.

Answer 1

The Correct Option is B

1. Given vectors: \[ \vec{u} = 3\hat{i} - \hat{j}, \quad \vec{v} = 2\hat{i} + \hat{j} - \lambda \hat{k} \]
2. Calculate the dot product $\vec{u} \cdot \vec{v}$: \[ \vec{u} \cdot \vec{v} = (3\hat{i} - \hat{j}) \cdot (2\hat{i} + \hat{j} - \lambda \hat{k}) = 6 + 1 = 7 \]
3. Calculate the magnitudes of $\vec{u}$ and $\vec{v}$: \[ |\vec{u}| = \sqrt{3^2 + (-1)^2} = \sqrt{10} \] \[ |\vec{v}| = \sqrt{2^2 + 1^2 + \lambda^2} = \sqrt{5 + \lambda^2} \]
4. Use the given cosine of the angle between $\vec{u}$ and $\vec{v}$: \[ \cos \theta = \frac{\vec{u} \cdot \vec{v}}{|\vec{u}| |\vec{v}|} = \frac{7}{\sqrt{10} \sqrt{5 + \lambda^2}} = \frac{\sqrt{5}}{2\sqrt{7}} \] \[ \frac{7}{\sqrt{10} \sqrt{5 + \lambda^2}} = \frac{\sqrt{5}}{2\sqrt{7}} \] \[ 7 \cdot 2\sqrt{7} = \sqrt{5} \cdot \sqrt{10} \cdot \sqrt{5 + \lambda^2} \] \[ 14\sqrt{7} = 5\sqrt{10} \sqrt{5 + \lambda^2} \] \[ 14\sqrt{7} = 5\sqrt{10} \sqrt{5 + \lambda^2} \] \[ \lambda^2 = 9 \implies \lambda = 3 \] 5. Decompose $\vec{v}$ into $\vec{v}_1$ and $\vec{v}_2$: \[ \vec{v} = \vec{v}_1 + \vec{v}_2 \] \[ |\vec{v}|^2 = |\vec{v}_1|^2 + |\vec{v}_2|^2 \] \[ |\vec{v}|^2 = 14 \] Therefore, the correct answer is (2) 14.

Answer 2

We have \(\vec{u}=3\hat{i}-\hat{j}\) and \(\vec{v}=2\hat{i}+\hat{j}-\lambda\hat{k}\) with angle \(\cos^{-1}\!\left(\dfrac{\sqrt{5}}{2\sqrt{7}}\right)\) between them. Write \(\vec{v}=\vec{v}_1+\vec{v}_2\) where \(\vec{v}_1\parallel\vec{u}\) and \(\vec{v}_2\perp\vec{u}\). We need \(|\vec{v}_1|^2+|\vec{v}_2|^2\).

Concept Used:

Use the cosine formula for the angle between vectors: \[ \cos\theta=\frac{\vec{u}\cdot\vec{v}}{|\vec{u}||\vec{v}|}. \] For an orthogonal decomposition \(\vec{v}=\vec{v}_1+\vec{v}_2\) with \(\vec{v}_1\parallel\vec{u}\) and \(\vec{v}_2\perp\vec{u}\), we have Pythagoras: \[ |\vec{v}_1|^2+|\vec{v}_2|^2=|\vec{v}|^2. \]

Step-by-Step Solution:

Step 1: Compute \(\lambda\) from the angle condition.

\[ \vec{u}\cdot\vec{v}=3\cdot2+(-1)\cdot1=5,\quad |\vec{u}|=\sqrt{3^2+(-1)^2}=\sqrt{10},\quad |\vec{v}|=\sqrt{2^2+1^2+\lambda^2}=\sqrt{5+\lambda^2}. \] \[ \frac{5}{\sqrt{10}\,\sqrt{5+\lambda^2}}=\frac{\sqrt{5}}{2\sqrt{7}} \ \Rightarrow\ \frac{5}{10(5+\lambda^2)^{1/2}}=\frac{\sqrt{5}}{2\sqrt{7}} \ \Rightarrow\ \frac{5}{2(5+\lambda^2)}=\frac{5}{28} \] \[ \Rightarrow\ 2(5+\lambda^2)=28\ \Rightarrow\ \lambda^2=9\ \Rightarrow\ \lambda=3\;(\lambda>0). \]

Step 2: Use orthogonal decomposition to get the required sum.

\[ |\vec{v}_1|^2+|\vec{v}_2|^2=|\vec{v}|^2=5+\lambda^2=5+9=14. \]

Final Computation & Result

\(|\vec{v}_1|^2+|\vec{v}_2|^2=14\).