MHT CET · Maths · Vector Algebra
The altitude of the parallelopiped, whose coterminus edges are the vectors \(\bar{a}=\hat{\mathrm{i}}+\hat{\mathrm{j}}+\hat{\mathrm{k}}, \overline{\mathrm{b}}=2 \hat{\mathrm{i}}+4 \hat{\mathrm{j}}-\hat{\mathrm{k}}, \overline{\mathrm{c}}=\hat{\mathrm{i}}+\hat{\mathrm{j}}+3 \hat{\mathrm{k}}\), where \(\bar{a}, \overline{\mathrm{~b}}\) are the sides of the base of parallelopiped, is
- A \(4/\sqrt{38}\)
- B \(2/\sqrt{38}\)
- C \(4/\sqrt{19}\)
- D \(2/\sqrt{19}\)
Answer & Solution
Correct Answer
(A) \(4/\sqrt{38}\)
Step-by-step Solution
Detailed explanation
\( V = |\bar{a} \cdot (\bar{b} \times \bar{c})| = |\begin{vmatrix} 1 & 1 & 1 \\ 2 & 4 & -1 \\ 1 & 1 & 3 \end{vmatrix}| = |1(12+1) - 1(6+1) + 1(2-4)| = |13-7-2| = 4 \) \( A = |\bar{a} \times \bar{b}| = |(\hat{\mathrm{i}}+\hat{\mathrm{j}}+\hat{\mathrm{k}}) \times (2\hat{\mathrm{i}}+4\hat{\mathrm{j}}-\hat{\mathrm{k}})| = |-5\hat{\mathrm{i}}+3\hat{\mathrm{j}}+2\hat{\mathrm{k}}| = \sqrt{(-5)^2+3^2+2^2} = \sqrt{25+9+4} = \sqrt{38} \)
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