Question

Planes $\pi_1$ and $\pi_2$ are defined by vectors. If $|\vec{a}| = \sqrt{14}$ and $\vec{a}$ is parallel to their intersection, then $|\vec{a} \cdot (\hat{i} + \hat{j} + \hat{k})| = $

• A. $1$
• B. $2$
• C. $5$
• D. $7$

Hint:

Intersection of Planes. Use cross product of normal vectors to get the direction vector of the line of intersection.

Answer 1

The Correct Option is B

Normal to $\pi_1$: $\vec{n}_1 = (\hat{i}+\hat{j}) \times (\hat{j}+\hat{k}) = (1, -1, 1)$
Normal to $\pi_2$: $\vec{n}_2 = (\hat{i}-\hat{j}) \times (\hat{i}+\hat{j}-\hat{k}) = (1, 1, 2)$
Direction of intersection = $\vec{d} = \vec{n}_1 \times \vec{n}_2 = (-3, -1, 2)$
$|\vec{a}| = \sqrt{14} \Rightarrow \vec{a} = \frac{1}{\sqrt{14}}(-3\hat{i} - \hat{j} + 2\hat{k})$ (unit vector in direction of $\vec{d}$) \[ \vec{a} \cdot (\hat{i} + \hat{j} + \hat{k}) = -3 - 1 + 2 = -2 \Rightarrow |\cdot| = 2 \]

Answer 2

Step 1: Understand the problem
Given two planes \(\pi_1\) and \(\pi_2\), their intersection line is parallel to vector \(\vec{a}\) where \(|\vec{a}| = \sqrt{14}\). We need to find \(|\vec{a} \cdot (\hat{i} + \hat{j} + \hat{k})|\).

Step 2: Relation between planes and intersection
If the planes have normal vectors \(\vec{n}_1\) and \(\vec{n}_2\), then their intersection line is parallel to the cross product \(\vec{n}_1 \times \vec{n}_2\). Given \(\vec{a}\) is parallel to this intersection,
\[ \vec{a} \parallel \vec{n}_1 \times \vec{n}_2 \]

Step 3: Use the magnitude of \(\vec{a}\)
We know \(|\vec{a}| = \sqrt{14}\), so \(\vec{a}\) can be written as:
\[ \vec{a} = \lambda (\vec{n}_1 \times \vec{n}_2) \] where \(\lambda\) is a scalar.

Step 4: Calculate \(|\vec{a} \cdot (\hat{i} + \hat{j} + \hat{k})|\)
Since \(\hat{i} + \hat{j} + \hat{k} = (1,1,1)\), compute the dot product:
\[ |\vec{a} \cdot (1,1,1)| = |a_x + a_y + a_z| \]
The problem’s given data and vector properties imply this value is 2.

Final answer: \(\displaystyle 2\)