मराठी

If → a , → B , → C Are Non-zero, Non-coplanar Vectors, Prove that the Following Vectors Are Coplanar: (1) 5 → → a +, − 3 → B + 5 → C and − 2 → a + 3 → B − 4 → C - Mathematics

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प्रश्न

If \[\vec{a}\], \[\vec{b}\], \[\vec{c}\] are non-zero, non-coplanar vectors, prove that the following vectors are coplanar:
(1) \[5 \vec{a} + 6 \vec{b} + 7 \vec{c,} 7 \vec{a} - 8 \vec{b} + 9 \vec{c}\text{ and }3 \vec{a} + 20 \vec{b} + 5 \vec{c}\]

(2) \[\vec{a} - 2 \vec{b} + 3 \vec{c} , - 3 \vec{b} + 5 \vec{c}\text{ and }- 2 \vec{a} + 3 \vec{b} - 4 \vec{c}\]
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उत्तर

(i) The three vectors are coplanar if one of them is expressible as a linear combination of the other two . Let \[5 \vec{a} + 6 \vec{b} + 7 \vec{c} = x \left( 7 \vec{a} - 8 \vec{b} + 9 c^\rightharpoonup \right) + y \left( 3 \vec{a} + 20 \vec{b} + 5 \vec{c} \right) . \]
\[ = \vec{a} \left( 7x + 3y \right) + \vec{b} \left( - 8x + 20y \right) + \vec{c} \left( 9x + 5y \right) .\]
\[\Rightarrow 7x + 3y = 5, - 8x + 20y = 6\text{ and }9x + 5y = 7 .\]
Solving first two of these equations, we get \[x = \frac{1}{2}, y = \frac{1}{2}\]. Clearly, these values of x and y satisfies the third equation.
Hence, the given vectors are coplanar.
(ii) The three vectors are coplanar if one of them is expressible as a linear combination of the other two.
Let \[\vec{a} - 2 \vec{b} + 3 \vec{c} = x ( - 3 \vec{b} + 5 \vec{c} ) + y ( - 2 \vec{a} + 3 \vec{b} - 4 \vec{c} ) . \]
\[ = \vec{a} ( - 2y) + \vec{b} ( - 3x + 3y) + \vec{c} (5x - 4y) .\]
\[\Rightarrow - 2y = 1, - 3x + 3y = - 2\text{ and }5x - 4y = 3\]
Solving first two of these equations, we get \[x = \frac{1}{6}, y = - \frac{1}{2}\] 
These values of x and y does not satisfy the third equation.
Hence, the given vectors are not coplanar.

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पाठ 23: Algebra of Vectors - Exercise 23.8 [पृष्ठ ६५]

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आरडी शर्मा Mathematics [English] Class 12
पाठ 23 Algebra of Vectors
Exercise 23.8 | Q 3 | पृष्ठ ६५

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