Understanding the Moment of a Force
The problem asks us to find the value of '\(\lambda\)' given a force vector '\(\vec{F}\)', a point of application A, a point B about which the moment is calculated, and the resulting moment vector '\(\vec{M}\)'.
The moment of a force '\(\vec{F}\)' about a point B, when the force is applied at point A, is defined as the cross product of the position vector '\(\vec{r}\)' (from B to A) and the force vector '\(\vec{F}\)'. Mathematically, this is expressed as:
\(\vec{M} = \vec{r} \times \vec{F}\)
where '\(\vec{r} = \vec{A} - \vec{B}\)'.
Calculating the Position Vector
First, we need to determine the position vector '\(\vec{r}\)' from point B\((-1, -2, 3)\) to point A\((1, 2, 5)\).
The coordinates of A are \((x_A, y_A, z_A) = (1, 2, 5)\).
The coordinates of B are \((x_B, y_B, z_B) = (-1, -2, 3)\).
The position vector '\(\vec{r}\)' is calculated as:
\(\vec{r} = (x_A - x_B)\hat{i} + (y_A - y_B)\hat{j} + (z_A - z_B)\hat{k}\) \(\vec{r} = (1 - (-1))\hat{i} + (2 - (-2))\hat{j} + (5 - 3)\hat{k}\) \(\vec{r} = (1 + 1)\hat{i} + (2 + 2)\hat{j} + (2)\hat{k}\) \(\vec{r} = 2\hat{i} + 4\hat{j} + 2\hat{k}\)Calculating the Moment Vector
Now, we calculate the moment vector '\(\vec{M}\)' using the cross product '\(\vec{r} \times \vec{F}\)'.
We have:
\(\vec{r} = 2\hat{i} + 4\hat{j} + 2\hat{k}\) \(\vec{F} = 2\hat{i} - \lambda\hat{j} + 5\hat{k}\)The cross product can be computed using a determinant:
$ \vec{M} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 4 & 2 \\ 2 & -\lambda & 5 \end{vmatrix} $Expanding the determinant:
\(\vec{M} = \hat{i}((4)(5) - (2)(-\lambda)) - \hat{j}((2)(5) - (2)(2)) + \hat{k}((2)(-\lambda) - (4)(2))\) \(\vec{M} = \hat{i}(20 + 2\lambda) - \hat{j}(10 - 4) + \hat{k}(-2\lambda - 8)\) \(\vec{M} = (20 + 2\lambda)\hat{i} - 6\hat{j} + (-2\lambda - 8)\hat{k}\)Finding the Value of Lambda (\(\lambda\))
We are given that the moment vector is '\(\vec{M}_{given} = 16\hat{i} - 6\hat{j} + 2\lambda\hat{k}\)'.
By comparing the calculated moment vector '\(\vec{M}\)' with the given moment vector '\(\vec{M}_{given}\)', we equate their corresponding components:
- X-component: '\((20 + 2\lambda) = 16\)'
- Y-component: '\(-6 = -6\)' (This confirms our calculation for the j-component)
- Z-component: '\((-2\lambda - 8) = 2\lambda\)'
Let's solve the equation from the X-component:
\(20 + 2\lambda = 16\) \(2\lambda = 16 - 20\) \(2\lambda = -4\) \(\lambda = \frac{-4}{2}\) \(\lambda = -2\)We can verify this using the Z-component equation:
\(-2\lambda - 8 = 2\lambda\) \(-8 = 2\lambda + 2\lambda\) \(-8 = 4\lambda\) \(\lambda = \frac{-8}{4}\) \(\lambda = -2\)Both component equations yield the same value for '\(\lambda\)'.
Conclusion
The value of '\(\lambda\)' that satisfies the condition for the moment of the force is -2.
Similar Questions
Let \(\vec{a}, \vec{b}, \vec{c}\) be three non-zero vectors such that \(\vec{a}\times \vec{b} = \vec{c} \) . Consider the following statements:
1. \(\vec a\) is unique if \(\vec b\) and \(\vec c\) are given
2. \(\vec c\) is unique if \(\vec a\) and \(\vec b\) are given
Which of the above statements is/are correct?
In a right angled triangle ABC, if the hypotenuse AC = p, then what is \(\overrightarrow {{\rm{AB}}} \cdot \overrightarrow {{\rm{AC}}} + \overrightarrow {{\rm{BC}}} \cdot \overrightarrow {{\rm{BA}}} + \overrightarrow {{\rm{CA}}} \cdot \overrightarrow {{\rm{CB}}} \) equal to?
What is \({\rm{\vec c}}\) equal to?
If \({\rm{\vec d}} = {\rm{x\hat i}} + {\rm{y\hat j}} + {\rm{z\hat k}}\) , then which of the following equations is/are correct?
1. y – x = 4
2. 2z – 3 = 0
Select the correct answer using the code given below:
What is \({\rm{\vec a}} \cdot {\rm{\vec b}} + {\rm{\vec b}} \cdot {\rm{\vec c}} + {\rm{\vec c}} \cdot {\rm{\vec a}}\) equal to?
What is the angle between \({\rm{\vec a}}\) and \({\rm{\vec b}}\) ?
- What is the fourth vertex \(D\)?
- Let \(\vec{a} = \hat{i} - \hat{j} + \hat{k}\) and \(\vec{b} = \hat{i} + 2\hat{j} - \hat{k}\). If \(\vec{a} \times (\vec{b} \times \vec{a}) = \alpha\hat{i} - \beta\hat{j} + \gamma\hat{k}\), then what is the value of \(\alpha + \beta + \gamma\)?
- Let \(\vec{p}=\vec{a}-\vec{b}\), \(\vec{q}=\vec{a}+\vec{b}\). If \(|\vec{a}|=|\vec{b}|= 2\) and \(\vec{a}\cdot\vec{b} = 2\), then what is the value of \(|\vec{p}\times\vec{q}|\)?
How many of the following can be a vector perpendicular to both the vectors \(2\hat{i} - \hat{j} + \hat{k}\) and \(\hat{i} + \hat{j} + 3\hat{k}\) ?
I. \(4\hat{i} + 5\hat{j} - 3\hat{k}\)
II. \(-8\hat{i} - 10\hat{j} + 6\hat{k}\)
III. \(\frac{1}{50}(-4\hat{i}-5\hat{j}+3\hat{k})\)
Select the correct answer.
Important Questions from Vector Algebra
Vector A̅ = ŷ.3 + ẑ.2 and B̅ = x̂.5 + ŷ.8 extend from the origin. Find A̅.B̅ Choose the correct answer.
The value of the cross product \(\left( {\overrightarrow a - \overrightarrow b } \right) \times \left( {\overrightarrow a + \overrightarrow b } \right)\) of two vectors \(\overrightarrow a - \overrightarrow b\) and \(\overrightarrow a + \overrightarrow b \) is:
If non - zero a, b, c are such that a + b + c = 0, then the value of \(\frac{a^2}{bc} + \frac{b^2}{ac} + \frac{c^2}{ab}\) is
Vector a = 3i + 2j – 6k, vector b = 4i – 3j + k, angle between above vectors is
Two forces F1 and F2 are used to pull a car, which met an accident. The angle between the two force is θ. Find the value of θ for the resultant force is equal to \(\sqrt{(F_1^2 + F_2^2)}\)
