Step 1: Understanding the direction vector of the line. The given line can be expressed as: \[ \vec{r} = \hat{i} + \hat{j} - \hat{k} + \lambda (3\hat{i} - \hat{j}) \] Here, the direction vector of the line is: \[ \vec{d} = 3\hat{i} - \hat{j} \]
Step 2: Condition for perpendicularity. A vector \( \vec{v} \) is perpendicular to \( \vec{d} \) if their dot product is zero: \[ \vec{v} \cdot \vec{d} = 0 \] Let \( \vec{v} = a\hat{i} + b\hat{j} + c\hat{k} \). The dot product is: \[ \vec{v} \cdot \vec{d} = (a\hat{i} + b\hat{j} + c\hat{k}) \cdot (3\hat{i} - \hat{j}) = 3a - b \] For perpendicularity: \[ 3a - b = 0 \quad \Rightarrow \quad b = 3a \quad \cdots (1) \]
Step 3: Using the options to find the correct vector. We substitute the options to check which satisfies \( b = 3a \): For \( \hat{i} + 3\hat{j} + 5\hat{k} \)
(Option \( B \)): \[ a = 1, \, b = 3, \, c = 5 \] Substituting into \( b = 3a \): \[ b = 3 \times 1 = 3 \quad \text{(True)} \] Therefore, this vector satisfies the condition.
Other options do not satisfy \( b = 3a \).
Conclusion: Thus, the required vector is \( \hat{i} + 3\hat{j} + 5\hat{k} \), which corresponds to option \( \mathbf{(B)} \).
Let $ \vec{a} = \hat{i} + 2\hat{j} + \hat{k} $, $ \vec{b} = 3\hat{i} - 3\hat{j} + 3\hat{k} $, $ \vec{c} = 2\hat{i} - \hat{j} + 2\hat{k} $ and $ \vec{d} $ be a vector such that $ \vec{b} \times \vec{d} = \vec{c} \times \vec{d} $ and $ \vec{a} \cdot \vec{d} = 4 $. Then $ |\vec{a} \times \vec{d}|^2 $ is equal to _______
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
The correct IUPAC name of \([ \text{Pt}(\text{NH}_3)_2\text{Cl}_2 ]^{2+} \) is: