KCET · Maths · Vector Algebra
If \(\mathbf{a}=(1,2,3), \mathbf{b}=(2,-1,1), \mathbf{c}=(3,2,1)\) and \(\mathbf{a} \times(\mathbf{b} \times \mathbf{c})=\alpha \mathbf{a}+\beta \mathbf{b}+\gamma \mathbf{c}\), then
- A \(\alpha=1, \beta=10, \gamma=3\)
- B \(\alpha=0, \beta=10, \gamma=-3\)
- C \(\alpha+\beta+\gamma=8\)
- D \(\alpha=\beta=\gamma=0\)
Answer & Solution
Correct Answer
(B) \(\alpha=0, \beta=10, \gamma=-3\)
Step-by-step Solution
Detailed explanation
Given,
\(a=(1,2,3)=i+2 j+3 k\)
\(b=(2,-1,1)=2 i-j+k\)
\(\begin{array}{ll}\text { and } & c=(3,2,1)=3 i+2 j+k \\ \text { Now, } & a \times(b \times c)=\alpha a+\beta b+\gamma c \\ \Rightarrow & (a \cdot c) b-(a \cdot b) c=\alpha a+\beta b+\gamma c \\ \Rightarrow & \{(i+2 j+3 k) \cdot(3 i+2 j+k)\} b \\ & \quad-\{(i+2 j+3 k) \cdot(2 i-j+k)\} c\end{array}\)
\(\begin{aligned}
&=\alpha a+\beta b+\gamma c \\
\Rightarrow &(3+4+3) b-(2-2+3) c=\alpha a+\beta b+\gamma c \\
\Rightarrow & 0 a+10 b-3 c=\alpha a+\beta b+\gamma c
\end{aligned}\)
On comparing, we get
\(\alpha=0, \beta=10 \text { and } \gamma=-3\)
\(a=(1,2,3)=i+2 j+3 k\)
\(b=(2,-1,1)=2 i-j+k\)
\(\begin{array}{ll}\text { and } & c=(3,2,1)=3 i+2 j+k \\ \text { Now, } & a \times(b \times c)=\alpha a+\beta b+\gamma c \\ \Rightarrow & (a \cdot c) b-(a \cdot b) c=\alpha a+\beta b+\gamma c \\ \Rightarrow & \{(i+2 j+3 k) \cdot(3 i+2 j+k)\} b \\ & \quad-\{(i+2 j+3 k) \cdot(2 i-j+k)\} c\end{array}\)
\(\begin{aligned}
&=\alpha a+\beta b+\gamma c \\
\Rightarrow &(3+4+3) b-(2-2+3) c=\alpha a+\beta b+\gamma c \\
\Rightarrow & 0 a+10 b-3 c=\alpha a+\beta b+\gamma c
\end{aligned}\)
On comparing, we get
\(\alpha=0, \beta=10 \text { and } \gamma=-3\)
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