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Question: Consider a vector equation $k\overrightarrow{x} + (\overrightarrow{x} \cdot \overrightarrow{a}) \ove...

Consider a vector equation kx→+(x→⋅a→)b→=c→,k≠0.k≠−b→⋅a→k\overrightarrow{x} + (\overrightarrow{x} \cdot \overrightarrow{a}) \overrightarrow{b} = \overrightarrow{c}, k \neq 0. k \neq -\overrightarrow{b} \cdot \overrightarrow{a}. Then x→\overrightarrow{x} is equal to

A

1k(((a→⋅b→)c→−c→⋅a→k+a→⋅b→a→)\frac{1}{k}(((\overrightarrow{a} \cdot \overrightarrow{b}) \overrightarrow{c} - \frac{\overrightarrow{c} \cdot \overrightarrow{a}}{k + \overrightarrow{a} \cdot \overrightarrow{b}}\overrightarrow{a})

B

1k(c→+c→⋅a→k+b→⋅a→b→)\frac{1}{k}(\overrightarrow{c} + \frac{\overrightarrow{c} \cdot \overrightarrow{a}}{k + \overrightarrow{b} \cdot \overrightarrow{a}}\overrightarrow{b})

C

1k(((a→⋅b→)c→+c→⋅a→k+a→⋅b→a→)\frac{1}{k}(((\overrightarrow{a} \cdot \overrightarrow{b}) \overrightarrow{c} + \frac{\overrightarrow{c} \cdot \overrightarrow{a}}{k + \overrightarrow{a} \cdot \overrightarrow{b}}\overrightarrow{a})

D

1k(c→−c→⋅a→k+b→⋅a→b→)\frac{1}{k}(\overrightarrow{c} - \frac{\overrightarrow{c} \cdot \overrightarrow{a}}{k + \overrightarrow{b} \cdot \overrightarrow{a}}\overrightarrow{b})

Answer

D

Explanation

Solution

Let the given vector equation be kx→+(x→⋅a→)b→=c→k\overrightarrow{x} + (\overrightarrow{x} \cdot \overrightarrow{a}) \overrightarrow{b} = \overrightarrow{c}. Let S=x→⋅a→S = \overrightarrow{x} \cdot \overrightarrow{a}. Since x→\overrightarrow{x} and a→\overrightarrow{a} are vectors, SS is a scalar. The equation becomes kx→+Sb→=c→k\overrightarrow{x} + S\overrightarrow{b} = \overrightarrow{c}. Rearranging to solve for x→\overrightarrow{x}: kx→=c→−Sb→k\overrightarrow{x} = \overrightarrow{c} - S\overrightarrow{b} x→=1k(c→−Sb→)\overrightarrow{x} = \frac{1}{k}(\overrightarrow{c} - S\overrightarrow{b})

Now, we find the scalar SS by taking the dot product of x→\overrightarrow{x} with a→\overrightarrow{a}: S=x→⋅a→S = \overrightarrow{x} \cdot \overrightarrow{a} Substitute the expression for x→\overrightarrow{x}: S=(1k(c→−Sb→))⋅a→S = \left(\frac{1}{k}(\overrightarrow{c} - S\overrightarrow{b})\right) \cdot \overrightarrow{a} Using the linearity property of the dot product: S=1k(c→⋅a→−(Sb→)⋅a→)S = \frac{1}{k} (\overrightarrow{c} \cdot \overrightarrow{a} - (S\overrightarrow{b}) \cdot \overrightarrow{a}) S=1k(c→⋅a→−S(b→⋅a→))S = \frac{1}{k} (\overrightarrow{c} \cdot \overrightarrow{a} - S(\overrightarrow{b} \cdot \overrightarrow{a})) Multiply both sides by kk: kS=c→⋅a→−S(b→⋅a→)kS = \overrightarrow{c} \cdot \overrightarrow{a} - S(\overrightarrow{b} \cdot \overrightarrow{a}) Rearrange to solve for SS: kS+S(b→⋅a→)=c→⋅a→kS + S(\overrightarrow{b} \cdot \overrightarrow{a}) = \overrightarrow{c} \cdot \overrightarrow{a} S(k+b→⋅a→)=c→⋅a→S(k + \overrightarrow{b} \cdot \overrightarrow{a}) = \overrightarrow{c} \cdot \overrightarrow{a} Given that k≠−b→⋅a→k \neq -\overrightarrow{b} \cdot \overrightarrow{a}, the term (k+b→⋅a→)(k + \overrightarrow{b} \cdot \overrightarrow{a}) is non-zero, so we can divide by it: S=c→⋅a→k+b→⋅a→S = \frac{\overrightarrow{c} \cdot \overrightarrow{a}}{k + \overrightarrow{b} \cdot \overrightarrow{a}}

Substitute this value of SS back into the expression for x→\overrightarrow{x}: x→=1k(c→−Sb→)\overrightarrow{x} = \frac{1}{k}(\overrightarrow{c} - S\overrightarrow{b}) x→=1k(c→−c→⋅a→k+b→⋅a→b→)\overrightarrow{x} = \frac{1}{k}\left(\overrightarrow{c} - \frac{\overrightarrow{c} \cdot \overrightarrow{a}}{k + \overrightarrow{b} \cdot \overrightarrow{a}}\overrightarrow{b}\right)