Question:medium

The magnitude of a vector which is orthogonal to the vector $\hat{i} + \hat{j} + \hat{k}$ and is coplanar with the vectors $\hat{i} + \hat{j} + 2\hat{k}$ and $\hat{i} + 2\hat{j} + \hat{k}$ is

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A vector coplanar with $B, C$ and perpendicular to $A$ follows the direction $(\bar{b} \times \bar{c}) \times \bar{a}$.
Updated On: May 14, 2026
  • $\sqrt{2}$
  • $4\sqrt{2}$
  • $4$
  • $2\sqrt{3}$
Show Solution

The Correct Option is A

Solution and Explanation

Step 1: Understanding the Concept:
A vector coplanar with vectors $\vec{b}$ and $\vec{c}$ can be expressed as a linear combination of them, or it lies in the plane defined by them.
A vector orthogonal to vector $\vec{a}$ has a dot product of zero with $\vec{a}$.
The vector that satisfies both conditions (coplanar with $\vec{b}, \vec{c}$ and orthogonal to $\vec{a}$) is parallel to the vector triple product $\vec{a} \times (\vec{b} \times \vec{c})$. Given the specific options, the question implicitly asks for the magnitude of this exact vector construction without any arbitrary scaling factor.
Step 2: Key Formula or Approach:
Let $\vec{a} = \hat{i} + \hat{j} + \hat{k}$.
Let $\vec{b} = \hat{i} + \hat{j} + 2\hat{k}$.
Let $\vec{c} = \hat{i} + 2\hat{j} + \hat{k}$.
The required vector is $\vec{v} = \vec{a} \times (\vec{b} \times \vec{c})$.
Vector Triple Product expansion: $\vec{a} \times (\vec{b} \times \vec{c}) = (\vec{a} \cdot \vec{c})\vec{b} - (\vec{a} \cdot \vec{b})\vec{c}$.
Step 3: Detailed Explanation:
Let's calculate the dot products first:
\[ \vec{a} \cdot \vec{c} = (\hat{i} + \hat{j} + \hat{k}) \cdot (\hat{i} + 2\hat{j} + \hat{k}) = (1)(1) + (1)(2) + (1)(1) = 1 + 2 + 1 = 4 \]
\[ \vec{a} \cdot \vec{b} = (\hat{i} + \hat{j} + \hat{k}) \cdot (\hat{i} + \hat{j} + 2\hat{k}) = (1)(1) + (1)(1) + (1)(2) = 1 + 1 + 2 = 4 \]
Now substitute these into the expansion formula to find $\vec{v}$:
\[ \vec{v} = (\vec{a} \cdot \vec{c})\vec{b} - (\vec{a} \cdot \vec{b})\vec{c} \]
\[ \vec{v} = 4\vec{b} - 4\vec{c} = 4(\vec{b} - \vec{c}) \]
Calculate $\vec{b} - \vec{c}$:
\[ \vec{b} - \vec{c} = (\hat{i} + \hat{j} + 2\hat{k}) - (\hat{i} + 2\hat{j} + \hat{k}) \]
\[ \vec{b} - \vec{c} = (1 - 1)\hat{i} + (1 - 2)\hat{j} + (2 - 1)\hat{k} = 0\hat{i} - \hat{j} + \hat{k} = -\hat{j} + \hat{k} \]
So, the required vector is:
\[ \vec{v} = 4(-\hat{j} + \hat{k}) = -4\hat{j} + 4\hat{k} \]
Now, find its magnitude:
\[ |\vec{v}| = \sqrt{0^2 + (-4)^2 + 4^2} = \sqrt{16 + 16} = \sqrt{32} \]
\[ |\vec{v}| = \sqrt{16 \times 2} = 4\sqrt{2} \]
Step 4: Final Answer:
The magnitude of the vector is $4\sqrt{2}$.
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