Question:
If the volume of a parallelopiped, whose coterminus edges are given by the vectors \(\vec{a}=\hat{i}+\hat{j}+n\hat{k}\), \(\vec{b}=2\hat{i}+4\hat{j}-n\hat{k}\) and \(\vec{c}=\hat{i}+n\hat{j}+3\hat{k}\) \((n\ge 0)\), is 158 cu. units, then :
If the volume of a parallelopiped, whose coterminus edges are given by the vectors \(\vec{a}=\hat{i}+\hat{j}+n\hat{k}\), \(\vec{b}=2\hat{i}+4\hat{j}-n\hat{k}\) and \(\vec{c}=\hat{i}+n\hat{j}+3\hat{k}\) \((n\ge 0)\), is 158 cu. units, then :
Show Hint
Volume is the absolute value of the scalar triple product (a 3x3 determinant of the coefficients); solve for \(n\) and test each option.
Updated On: Nov 5, 2026
- \(n=7\)
- \(\vec{b}\cdot\vec{c}=10\)
- \(\vec{a}\cdot\vec{c}=17\)
- \(n=9\)
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The Correct Option is B
Solution and Explanation
Volume \(=|\vec{a}\cdot(\vec{b}\times\vec{c})|\), the absolute value of the determinant with rows \((1,1,n)\), \((2,4,-n)\), \((1,n,3)\).
Expanding: \(1(12+n^2)-1(6+n)+n(2n-4)=3n^2-5n+6\).
Setting \(3n^2-5n+6=158\): \(3n^2-5n-152=0\), so \(n=\frac{5\pm 43}{6}\), giving \(n=8\) (the other root \(-\frac{19}{3}\) is rejected as \(n\ge0\)). (The value \(-158\) gives \(3n^2-5n+164=0\), which has no real root.)
Check the options with \(n=8\): (1) \(n=7\) is false; (4) \(n=9\) is false.
\(\vec{b}\cdot\vec{c}=2+4n-3n=2+n=10\), true. \(\vec{a}\cdot\vec{c}=1+n+3n=1+4n=33\ne17\).
Answer: (2) \(\vec{b}\cdot\vec{c}=10\)
Expanding: \(1(12+n^2)-1(6+n)+n(2n-4)=3n^2-5n+6\).
Setting \(3n^2-5n+6=158\): \(3n^2-5n-152=0\), so \(n=\frac{5\pm 43}{6}\), giving \(n=8\) (the other root \(-\frac{19}{3}\) is rejected as \(n\ge0\)). (The value \(-158\) gives \(3n^2-5n+164=0\), which has no real root.)
Check the options with \(n=8\): (1) \(n=7\) is false; (4) \(n=9\) is false.
\(\vec{b}\cdot\vec{c}=2+4n-3n=2+n=10\), true. \(\vec{a}\cdot\vec{c}=1+n+3n=1+4n=33\ne17\).
Answer: (2) \(\vec{b}\cdot\vec{c}=10\)
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