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Suppose \({{\mathop{\rm v}\nolimits} _1},{{\mathop{\rm v}\nolimits} _2},{{\mathop{\rm v}\nolimits} _3}\) are distinct points on one line in \({\mathbb{R}^3}\). The line need not pass through the origin. Show that \(\left\{ {{{\mathop{\rm v}\nolimits} _1},{{\mathop{\rm v}\nolimits} _2},{{\mathop{\rm v}\nolimits} _3}} \right\}\) is linearly dependent.

Short Answer

Expert verified

The set \(\left\{ {{{\mathop{\rm v}\nolimits} _1},{{\mathop{\rm v}\nolimits} _2},{{\mathop{\rm v}\nolimits} _3}} \right\}\) is linearly dependent.

Step by step solution

01

Prove that the line passes through the origin

Consider \(M\) is the line that passes through the origin and is parallel to the line passing through \({{\mathop{\rm v}\nolimits} _1},{{\mathop{\rm v}\nolimits} _2},\) and \({{\mathop{\rm v}\nolimits} _3}\). Thus, both \({{\mathop{\rm v}\nolimits} _2} - {{\mathop{\rm v}\nolimits} _1}\) and \({{\mathop{\rm v}\nolimits} _3} - {{\mathop{\rm v}\nolimits} _1}\) are on \(M\).

02

Show that the set is linearly dependent

Since one of the two vectors is a multiple of the other, \({{\mathop{\rm v}\nolimits} _2} - {{\mathop{\rm v}\nolimits} _1} = k\left( {{{\mathop{\rm v}\nolimits} _3} - {{\mathop{\rm v}\nolimits} _1}} \right)\). The linear dependence relation produced by this equation resolves to \(\left( {k - 1} \right){{\mathop{\rm v}\nolimits} _1} + {{\mathop{\rm v}\nolimits} _2} - k{{\mathop{\rm v}\nolimits} _3} = 0\).

Thus, the set \(\left\{ {{{\mathop{\rm v}\nolimits} _1},{{\mathop{\rm v}\nolimits} _2},{{\mathop{\rm v}\nolimits} _3}} \right\}\) is linearly dependent.

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