The altitude of a parallelepiped whose three coterminous edges are vectors A⃗=i^+j^+k^‚B⃗=2i^+4j^−k^andC⃗=1^+j^+3k^,withA⃗andB⃗ \vec{A} = \hat{i} + \hat{j} + \hat{k} ‚ \vec{B} = 2 \hat{i} + 4 \hat{j} − \hat{k} \quad ext{and}\quad \vec{C} = \hat{1} + \hat{j} + 3 \hat{k} , w i t h \vec{A} \quad ext{and}\quad \vec{B} as sides of the base of the parallelepiped is-

Question

The altitude of a parallelepiped whose three coterminous edges are vectors A⃗=i^+j^+k^‚B⃗=2i^+4j^−k^andC⃗=1^+j^+3k^,withA⃗andB⃗ \vec{A} = \hat{i} + \hat{j} + \hat{k} ‚ \vec{B} = 2 \hat{i} + 4 \hat{j} − \hat{k} \quad	ext{and}\quad \vec{C} = \hat{1} + \hat{j} + 3 \hat{k} , w i t h \vec{A} \quad	ext{and}\quad \vec{B} as sides of the base of the parallelepiped is-

Accepted Answer

গ. 2√3819 \frac{2 √ 38}{19}

ব্যাখ্যা: সামান্তরিকীয় ঘনকের উচ্চতা $h = \frac{V}{|\vec{A} 	imes \vec{B}|}$। এখানে আয়তন $V = |\vec{A} \cdot (\vec{B} 	imes \vec{C})| = \begin{vmatrix} 1 & 1 & 1 \ 2 & 4 & -1 \ 1 & 1 & 3 \end{vmatrix} = 1(12 - (-1)) - 1(6 - (-1)) + 1(2 - 4) = 13 - 7 - 2 = 4$। ভূমির ক্ষেত্রফল $|\vec{A} 	imes \vec{B}| = |(-1-4)\hat{i} - (-1-2)\hat{j} + (4-2)\hat{k}| = |-5\hat{i} + 3\hat{j} + 2\hat{k}| = \sqrt{(-5)^2 + 3^2 + 2^2} = \sqrt{25+9+4} = \sqrt{38}$। অতএব, উচ্চতা $h = \frac{4}{\sqrt{38}} = \frac{2\sqrt{38}}{19}$।

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