Let vectors $\mathbf{a}, \mathbf{b}, \mathbf{c}$ be such that $$ \mathbf{a} = \hat{i} + 2\hat{j} - \hat{k}, \quad \mathbf{b} = 2\hat{i} - \hat{j} + \hat{k}, \quad \mathbf{c} = \hat{i} + \hat{j} + \hat{k} $$ Then the volume of the parallelepiped formed by these vectors is:

Question

Let $ \vec{a}, \vec{b}, \vec{c} $ be vectors of equal magnitude such that the angle between $ \vec{a} $ and $ \vec{b} $ is $ \alpha $, the angle between $ \vec{b} $ and $ \vec{c} $ is $ \beta $, and the angle between $ \vec{c} $ and $ \vec{a} $ is $ \gamma $. Then the minimum value of $ \cos \alpha + \cos \beta + \cos \gamma $ is:

Answer 1

The volume of the parallelepiped defined by vectors a, b, and c is determined by their scalar triple product: Volume = a ⋅ (<strong>b> × c).

Given vectors:

a = î + 2ĵ - k̂,
b = 2î - ĵ + k̂,
c = î + ĵ + k̂.

Calculate the cross product b × c using the formula:

u × v = (u₂v₃ - u₃v₂)î + (u₃v₁ - u₁v₃)ĵ + (u₁v₂ - u₂v₁)k̂.

For b and c:

b × c = (−1×1 − 1×1)î + (1×1 − 2×1)ĵ + (2×1 − (−1)×1)k̂

= (−1 − 1)î + (1 − 2)ĵ + (2 + 1)k̂

= −2î − 1ĵ + 3k̂.

Next, compute the dot product of a with the result of b × c:

d ⋅ v = d₁v₁ + d₂v₂ + d₃v₃.

Using a = î + 2ĵ - k̂ and b × c = −2î − 1ĵ + 3k̂:

a ⋅ (<strong>b> × c) = 1×(−2) + 2×(−1) + (−1)×3

= −2 − 2 − 3

= −7.

The volume of the parallelepiped is therefore -7.

Let vectors $\mathbf{a}, \mathbf{b}, \mathbf{c}$ be such that $$ \mathbf{a} = \hat{i} + 2\hat{j} - \hat{k}, \quad \mathbf{b} = 2\hat{i} - \hat{j} + \hat{k}, \quad \mathbf{c} = \hat{i} + \hat{j} + \hat{k} $$ Then the volume of the parallelepiped formed by these vectors is:

Show Hint

The Correct Option is A

Solution and Explanation

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