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Find the Value of λ So that the Following Vector is Coplanar: → a = 2 ^ I − ^ J + ^ K , → B =

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प्रश्न

Find the value of λ so that the following vector is coplanar:

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

योग
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उत्तर

Given: 

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

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

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

\[\text { We know that vectors } \vec{a} , \vec{b} , \vec{c}\text {  are coplanar iff } \left[ \vec{a} \vec{b} \vec{c} \right] = 0 . \]

\[\text { It is given that} \vec{a} , \vec{b} , \vec{c} \text { are coplanar } . \]

\[ \therefore \left[ \vec{a} \vec{b} \vec{c} \right] = 0\]

\[ \Rightarrow \begin{vmatrix}2 & - 1 & 1 \\ 1 & 2 & - 3 \\ \lambda & \lambda & 5\end{vmatrix} = 0 \]

\[ \Rightarrow 2\left( 10 + 3\lambda \right) + 1\left( 5 + 3\lambda \right) + 1\left( \lambda - 2\lambda \right) = 0\]

\[ \Rightarrow 8\lambda + 25 = 0 \]

\[ \Rightarrow \lambda = - \frac{25}{8}\]

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

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आर.डी. शर्मा Mathematics Volume 1 and 2 [English] Class 12
अध्याय 25 Scalar Triple Product
Exercise 26.1 | Q 5.2 | पृष्ठ १६
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