Question

Let $\vec{a}\times(2\hat{i}+3\hat{j}+4\hat{k})=(2\hat{i}+3\hat{j}+4\hat{k})\times\vec{b}$. If $|\vec{a}+\vec{b}|=\sqrt{29}$, then $\vec{a}+\vec{b} = $ ________.

• A. $(2\hat{i}+3\hat{j}-4\hat{k})$
• B. $-(2\hat{i}+3\hat{j}-4\hat{k})$
• C. $\pm(2\hat{i}+3\hat{j}+4\hat{k})$
• D. $\pm(2\hat{i}-3\hat{j}+4\hat{k})$
• E. $\pm\sqrt{29}(2\hat{i}+3\hat{j}+4\hat{k})$

Hint:

$\vec{A} \times \vec{B} = 0 \iff \vec{A}$ is parallel to $\vec{B}$.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept
The problem involves properties of the vector cross product. We are given an equation relating the cross products of three vectors and need to find a possible value for the sum of two of them.
Step 2: Key Formula or Approach
We will use the anti-commutative property of the cross product: \(\vec{x} \times \vec{y} = -(\vec{y} \times \vec{x})\).
Let \(\vec{c} = 2\hat{i} + 3\hat{j} + 4\hat{k}\). The given equation is \(\vec{a} \times \vec{c} = \vec{c} \times \vec{b}\).
We rearrange this equation to group \(\vec{a}\) and \(\vec{b}\) together.
Step 3: Detailed Explanation
1. Rearrange the given vector equation.
The equation is \(\vec{a} \times \vec{c} = \vec{c} \times \vec{b}\).
Using the anti-commutative property, \(\vec{c} \times \vec{b} = -(\vec{b} \times \vec{c})\).
So, \(\vec{a} \times \vec{c} = -(\vec{b} \times \vec{c})\).
Move all terms to one side:
\[ \vec{a} \times \vec{c} + \vec{b} \times \vec{c} = \vec{0} \] 2. Use the distributive property of the cross product.
\[ (\vec{a} + \vec{b}) \times \vec{c} = \vec{0} \] 3. Interpret the result.
The cross product of two non-zero vectors is the zero vector if and only if the two vectors are parallel (or one of them is the zero vector).
This means the vector \((\vec{a} + \vec{b})\) is parallel to the vector \(\vec{c}\).
Two vectors are parallel if one is a scalar multiple of the other. So, we can write:
\[ \vec{a} + \vec{b} = \lambda \vec{c} \] where \(\lambda\) is some scalar. Substituting the expression for \(\vec{c}\):
\[ \vec{a} + \vec{b} = \lambda (2\hat{i} + 3\hat{j} + 4\hat{k}) \] 4. Use the magnitude condition to find \(\lambda\).
We are given \(|\vec{a} + \vec{b}| = \sqrt{29}\).
Let's find the magnitude of \(\lambda \vec{c}\):
\[ |\lambda \vec{c}| = |\lambda| |\vec{c}| = |\lambda| \sqrt{2^2 + 3^2 + 4^2} = |\lambda| \sqrt{4 + 9 + 16} = |\lambda| \sqrt{29} \] Now we equate the magnitudes:
\[ |\lambda| \sqrt{29} = \sqrt{29} \] \[ |\lambda| = 1 \] This implies that \(\lambda = 1\) or \(\lambda = -1\).
5. Find the possible values for \(\vec{a} + \vec{b}\).
- If \(\lambda = 1\), then \(\vec{a} + \vec{b} = 1 \cdot (2\hat{i} + 3\hat{j} + 4\hat{k}) = 2\hat{i} + 3\hat{j} + 4\hat{k}\).
- If \(\lambda = -1\), then \(\vec{a} + \vec{b} = -1 \cdot (2\hat{i} + 3\hat{j} + 4\hat{k}) = -(2\hat{i} + 3\hat{j} + 4\hat{k})\).
Combining these, we get \(\vec{a} + \vec{b} = \pm(2\hat{i} + 3\hat{j} + 4\hat{k})\).
Step 4: Final Answer
The value of \(\vec{a} + \vec{b}\) is \(\pm(2\hat{i} + 3\hat{j} + 4\hat{k})\).