Let a vector \( \overrightarrow c \) be coplanar with the vectors \( \overrightarrow a = - \widehat i + \widehat j + \widehat k \) and \( \overrightarrow b = 2\widehat i + \widehat j - \widehat k \). If the vector \( \overrightarrow c \) also satisfies the conditions \( \overrightarrow c \,.\,\left[ {\left( {\overrightarrow a + \overrightarrow b } \right) \times \left( {\overrightarrow a \times \overrightarrow b } \right)} \right] = - 42 \) and \( \left( {\overrightarrow c \times \left( {\overrightarrow a - \overrightarrow b } \right)} \right)\,.\,\widehat k = 3 \), then the value of \( |\overrightarrow c {|^2} \) is equal to :
Let a vector \( \overrightarrow c \) be coplanar with the vectors \( \overrightarrow a = - \widehat i + \widehat j + \widehat k \) and \( \overrightarrow b = 2\widehat i + \widehat j - \widehat k \). If the vector \( \overrightarrow c \) also satisfies the conditions \( \overrightarrow c \,.\,\left[ {\left( {\overrightarrow a + \overrightarrow b } \right) \times \left( {\overrightarrow a \times \overrightarrow b } \right)} \right] = - 42 \) and \( \left( {\overrightarrow c \times \left( {\overrightarrow a - \overrightarrow b } \right)} \right)\,.\,\widehat k = 3 \), then the value of \( |\overrightarrow c {|^2} \) is equal to :
Show Hint
- 24
- 29
- 35
- 42
The Correct Option is C
Solution and Explanation
Step 2: \( (\vec a+\vec b)\times(\vec a\times\vec b)=(1,2,0)\times(-2,1,-3)=(-6,\,3,\,5) \).
Step 3: Coplanar: \( \vec c=\lambda\vec a+\mu\vec b=(-\lambda+2\mu,\ \lambda+\mu,\ \lambda-\mu) \).
Step 4: \( \vec c\cdot(-6,3,5)=6\lambda-12\mu+3\lambda+3\mu+5\lambda-5\mu=14(\lambda-\mu)=-42 \Rightarrow \lambda-\mu=-3 \).
Step 5: \( \vec a-\vec b=(-3,0,2) \). The k-component of \( \vec c\times(\vec a-\vec b) \) is \( c_x\cdot 0-c_y\cdot(-3)=3c_y=3 \Rightarrow c_y=1 \Rightarrow \lambda+\mu=1 \).
Step 6: Solving, \( \lambda=-1,\ \mu=2 \), so \( \vec c=(5,1,-3) \) and \( |\vec c|^2=25+1+9=35 \).
Answer: (C) 35
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