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If a, b and c are the three vectors mutu...

If a, b and c are the three vectors mutually perpendicular to each other to form a right handed system and `|a|=1, |b|=3 and |c|=5`, then `[a-2b b-3c c-4a]` is equal to

A

0

B

`-24`

C

3600

D

`-215`

Text Solution

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The correct Answer is:
To solve the problem, we need to find the scalar triple product of the vectors \( \mathbf{A} - 2\mathbf{B} \), \( \mathbf{B} - 3\mathbf{C} \), and \( \mathbf{C} - 4\mathbf{A} \). Let's denote these vectors as follows: - \( \mathbf{u} = \mathbf{A} - 2\mathbf{B} \) - \( \mathbf{v} = \mathbf{B} - 3\mathbf{C} \) - \( \mathbf{w} = \mathbf{C} - 4\mathbf{A} \) ### Step 1: Express the vectors in terms of their components Given that \( |\mathbf{A}| = 1 \), \( |\mathbf{B}| = 3 \), and \( |\mathbf{C}| = 5 \), we can represent these vectors in a right-handed coordinate system as: - \( \mathbf{A} = (1, 0, 0) \) - \( \mathbf{B} = (0, 3, 0) \) - \( \mathbf{C} = (0, 0, 5) \) ### Step 2: Calculate \( \mathbf{u} \), \( \mathbf{v} \), and \( \mathbf{w} \) Now, we can calculate each vector: - \( \mathbf{u} = \mathbf{A} - 2\mathbf{B} = (1, 0, 0) - 2(0, 3, 0) = (1, -6, 0) \) - \( \mathbf{v} = \mathbf{B} - 3\mathbf{C} = (0, 3, 0) - 3(0, 0, 5) = (0, 3, -15) \) - \( \mathbf{w} = \mathbf{C} - 4\mathbf{A} = (0, 0, 5) - 4(1, 0, 0) = (-4, 0, 5) \) ### Step 3: Compute the scalar triple product \( [\mathbf{u}, \mathbf{v}, \mathbf{w}] \) The scalar triple product can be calculated using the determinant of a matrix formed by the vectors: \[ [\mathbf{u}, \mathbf{v}, \mathbf{w}] = \begin{vmatrix} 1 & -6 & 0 \\ 0 & 3 & -15 \\ -4 & 0 & 5 \end{vmatrix} \] ### Step 4: Calculate the determinant To compute the determinant, we can use the formula for a 3x3 matrix: \[ \text{det} = a(ei-fh) - b(di-fg) + c(dh-eg) \] Substituting the values: \[ = 1(3 \cdot 5 - (-15) \cdot 0) - (-6)(0 \cdot 5 - (-15)(-4)) + 0(0 \cdot 0 - 3 \cdot (-4)) \] \[ = 1(15) + 6(60) + 0 \] \[ = 15 + 360 = 375 \] ### Final Answer Thus, the scalar triple product \( [\mathbf{A} - 2\mathbf{B}, \mathbf{B} - 3\mathbf{C}, \mathbf{C} - 4\mathbf{A}] \) is equal to \( 375 \).

To solve the problem, we need to find the scalar triple product of the vectors \( \mathbf{A} - 2\mathbf{B} \), \( \mathbf{B} - 3\mathbf{C} \), and \( \mathbf{C} - 4\mathbf{A} \). Let's denote these vectors as follows: - \( \mathbf{u} = \mathbf{A} - 2\mathbf{B} \) - \( \mathbf{v} = \mathbf{B} - 3\mathbf{C} \) - \( \mathbf{w} = \mathbf{C} - 4\mathbf{A} \) ### Step 1: Express the vectors in terms of their components Given that \( |\mathbf{A}| = 1 \), \( |\mathbf{B}| = 3 \), and \( |\mathbf{C}| = 5 \), we can represent these vectors in a right-handed coordinate system as: ...
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MHTCET PREVIOUS YEAR PAPERS AND PRACTICE PAPERS-VECTORS -( MHT CET Corner)
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