KCET · Physics · Current Electricity
If \( \vec{A}=2 \hat{i}+3 \hat{j}+8 \hat{k} \) is perpendicular to \( \vec{B}=4 \hat{j}-4 \hat{i}+\alpha \hat{k} \), then the value of ' \( \alpha \) ' is
- A \( \frac{1}{2} \)
- B \(-\frac{1}{2} \)
- C \( 11 \)
- D \( -1 \)
Answer & Solution
Correct Answer
(A) \( \frac{1}{2} \)
Step-by-step Solution
Detailed explanation
Given, \(\vec{A}=2 \hat{i}+3 \hat{j}+8 \hat{k}\) and \(\vec{B}=4 \hat{j}-4 \hat{i}+\alpha \hat{k}\)
Now, \(\vec{A} \cdot \vec{B}=|A||B| \cos \theta\)
Given, the two vectors are perpendicular. Therefore \(\theta=90^{\circ}\)
\(\Rightarrow \vec{A} \cdot \vec{B}=\left.|A|\right|_{B} \mid \cos 90^{\circ}\)
\(\Rightarrow \vec{A} \cdot \vec{B}=0\)
\(\Rightarrow(2 i+3 j+8 k) \cdot(4 j-4 i+\alpha k)=0\)
\(\Rightarrow 8-12+8 \alpha=0\)
\(\Rightarrow-4+8 \alpha=0\)
\(\Rightarrow \alpha=\frac{1}{2}\)
Now, \(\vec{A} \cdot \vec{B}=|A||B| \cos \theta\)
Given, the two vectors are perpendicular. Therefore \(\theta=90^{\circ}\)
\(\Rightarrow \vec{A} \cdot \vec{B}=\left.|A|\right|_{B} \mid \cos 90^{\circ}\)
\(\Rightarrow \vec{A} \cdot \vec{B}=0\)
\(\Rightarrow(2 i+3 j+8 k) \cdot(4 j-4 i+\alpha k)=0\)
\(\Rightarrow 8-12+8 \alpha=0\)
\(\Rightarrow-4+8 \alpha=0\)
\(\Rightarrow \alpha=\frac{1}{2}\)
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