मराठी

Show of the Following Triad of Vector is Coplanar: ^ a = ^ I − 2 ^ J + 3 ^ K , ^ B =

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प्रश्न

Show of the following triad of vector is coplanar:

\[\hat{a} = \hat{i} - 2 \hat {j} + 3 \hat {k} , \hat {b} = - 2 \hat {i} + 3 \hat {j} - 4 \hat { k}, \hat {c} = \hat { i} - 3 \hat { j} + 5 \hat { k }\]

बेरीज
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उत्तर

 Given: 

\[ \vec{a} = \hat { i} - 2 \hat{j} + 3 \hat{k} \]

\[ \vec{b} = - 2 \hat{i} + 3 \hat{j} - 4 \hat{k} \]

\[ \vec{c} = \hat{i} - 3 \hat{j} + 5\hat{ k} \]

\[\text { We know that three vectors }\vec{a} , \vec{b} , \vec{c} \text { are coplanar iff their scalar triple product is zero, i . e .} \left[ \vec{a} \vec{b} \vec{c} \right] = 0 . \]

Here,

\[\left[ \vec{a} \vec{b} \vec{c} \right] = \begin{vmatrix}1 & - 2 & 3 \\ - 2 & 3 & - 4 \\ 1 & - 3 & 5\end{vmatrix} = 1\left( 15 - 12 \right) + 2\left( - 10 + 4 \right) + 3\left( 6 - 3 \right) = 0\]

Hence, the given vectors are coplanar .

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पाठ 25: Scalar Triple Product - Exercise 26.1 [पृष्ठ १६]

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आर.डी. शर्मा Mathematics Volume 1 and 2 [English] Class 12
पाठ 25 Scalar Triple Product
Exercise 26.1 | Q 4.3 | पृष्ठ १६
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