KCET · Maths · Vector Algebra
If \(\overrightarrow{\mathbf{a}}+2 \overrightarrow{\mathbf{b}}+3 \overrightarrow{\mathbf{c}}=\overrightarrow{\mathbf{0}}\), then \(\overrightarrow{\mathbf{a}} \times \overrightarrow{\mathbf{b}}+\overrightarrow{\mathbf{b}} \times \overrightarrow{\mathbf{c}}+\overrightarrow{\mathbf{c}} \times \overrightarrow{\mathbf{a}}\) is equal to
- A \(2(\vec{b} \times \overrightarrow{\mathbf{c}})\)
- B \(3(\overrightarrow{\mathbf{c}} \times \overrightarrow{\mathbf{a}})\)
- C \(\overrightarrow{\mathbf{0}}\)
- D \(6(\overrightarrow{\mathbf{b}} \times \overrightarrow{\mathbf{c}})\)
Answer & Solution
Correct Answer
(D) \(6(\overrightarrow{\mathbf{b}} \times \overrightarrow{\mathbf{c}})\)
Step-by-step Solution
Detailed explanation
Given, \(\quad \overrightarrow{\mathbf{a}}+2 \overrightarrow{\mathbf{b}}+3 \overrightarrow{\mathbf{c}}=\overrightarrow{\mathbf{0}}
\quad \text{...(i)}\)
Taking cross product with \(\overrightarrow{\mathbf{b}}\), we get
\(\overrightarrow{\mathbf{a}} \times \overrightarrow{\mathbf{b}}+2 \overrightarrow{\mathbf{b}} \times \overrightarrow{\mathbf{b}}+3 \overrightarrow{\mathbf{c}} \times \overrightarrow{\mathbf{b}}=\overrightarrow{\mathbf{0}} \times \overrightarrow{\mathbf{b}}\)
\(\Rightarrow \quad \overrightarrow{\mathbf{a}} \times \overrightarrow{\mathbf{b}}=3 \overrightarrow{\mathbf{b}} \times \overrightarrow{\mathbf{c}}\)
Again taking cross product with \(\overrightarrow{\mathbf{c}}\) of Eq. (i), we get
\[
\overrightarrow{\mathbf{a}} \times \overrightarrow{\mathbf{c}}+2 \overrightarrow{\mathbf{b}} \times \overrightarrow{\mathbf{c}}+3 \overrightarrow{\mathbf{c}} \times \overrightarrow{\mathbf{c}}=\overrightarrow{\mathbf{0}} \times \overrightarrow{\mathbf{c}}
\]
\(\Rightarrow \quad \overrightarrow{\mathbf{c}} \times \overrightarrow{\mathbf{a}}=2 \overrightarrow{\mathbf{b}} \times \overrightarrow{\mathbf{c}}\)
\(\therefore \quad \overrightarrow{\mathbf{a}} \times \overrightarrow{\mathbf{b}}+\overrightarrow{\mathbf{b}} \times \overrightarrow{\mathbf{c}}+\overrightarrow{\mathbf{c}} \times \overrightarrow{\mathbf{a}}\)
\(=3 \overrightarrow{\mathbf{b}} \times \overrightarrow{\mathbf{c}}+\overrightarrow{\mathbf{b}} \times \overrightarrow{\mathbf{c}}+2 \overrightarrow{\mathbf{b}} \times \overrightarrow{\mathbf{c}}\)
\(=6(\overrightarrow{\mathbf{b}} \times \overrightarrow{\mathbf{c}})\)
\quad \text{...(i)}\)
Taking cross product with \(\overrightarrow{\mathbf{b}}\), we get
\(\overrightarrow{\mathbf{a}} \times \overrightarrow{\mathbf{b}}+2 \overrightarrow{\mathbf{b}} \times \overrightarrow{\mathbf{b}}+3 \overrightarrow{\mathbf{c}} \times \overrightarrow{\mathbf{b}}=\overrightarrow{\mathbf{0}} \times \overrightarrow{\mathbf{b}}\)
\(\Rightarrow \quad \overrightarrow{\mathbf{a}} \times \overrightarrow{\mathbf{b}}=3 \overrightarrow{\mathbf{b}} \times \overrightarrow{\mathbf{c}}\)
Again taking cross product with \(\overrightarrow{\mathbf{c}}\) of Eq. (i), we get
\[
\overrightarrow{\mathbf{a}} \times \overrightarrow{\mathbf{c}}+2 \overrightarrow{\mathbf{b}} \times \overrightarrow{\mathbf{c}}+3 \overrightarrow{\mathbf{c}} \times \overrightarrow{\mathbf{c}}=\overrightarrow{\mathbf{0}} \times \overrightarrow{\mathbf{c}}
\]
\(\Rightarrow \quad \overrightarrow{\mathbf{c}} \times \overrightarrow{\mathbf{a}}=2 \overrightarrow{\mathbf{b}} \times \overrightarrow{\mathbf{c}}\)
\(\therefore \quad \overrightarrow{\mathbf{a}} \times \overrightarrow{\mathbf{b}}+\overrightarrow{\mathbf{b}} \times \overrightarrow{\mathbf{c}}+\overrightarrow{\mathbf{c}} \times \overrightarrow{\mathbf{a}}\)
\(=3 \overrightarrow{\mathbf{b}} \times \overrightarrow{\mathbf{c}}+\overrightarrow{\mathbf{b}} \times \overrightarrow{\mathbf{c}}+2 \overrightarrow{\mathbf{b}} \times \overrightarrow{\mathbf{c}}\)
\(=6(\overrightarrow{\mathbf{b}} \times \overrightarrow{\mathbf{c}})\)
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