Consider the following for the next two (02) items that follow : Let \(\vec{a}=3 \hat{i}+3 \hat{j}+3 \hat{k} \) and \( \vec{c}=\hat{j}-\hat{k} \text {. Let } \vec{b}\) be such that \(\vec{a} \cdot \vec{b}=27 \) and \( \vec{a} \times \vec{b}=\overrightarrow{9 c}\)
What is \(\vec{b}\) equal to ?
Finding Vector b Using Dot and Cross Product Conditions
We are given two vectors, \(\vec{a} = 3 \hat{i} + 3 \hat{j} + 3 \hat{k}\) and \(\vec{c} = \hat{j} - \hat{k}\). We are also given two conditions involving a third vector \(\vec{b}\):
- The dot product of \(\vec{a}\) and \(\vec{b}\) is 27: \(\vec{a} \cdot \vec{b} = 27\)
- The cross product of \(\vec{a}\) and \(\vec{b}\) is 9 times vector \(\vec{c}\): \(\vec{a} \times \vec{b} = 9\vec{c}\)
Our goal is to find the components of the vector \(\vec{b}\).
Setting up the Vector \(\vec{b}\)
Let the unknown vector \(\vec{b}\) be represented by its components in the Cartesian coordinate system:
\(\vec{b} = x\hat{i} + y\hat{j} + z\hat{k}\)
where x, y, and z are unknown scalar values that we need to determine.
Using the Dot Product Condition (\(\vec{a} \cdot \vec{b} = 27\))
The dot product of two vectors is the sum of the products of their corresponding components. Using the given condition:
\((3 \hat{i} + 3 \hat{j} + 3 \hat{k}) \cdot (x\hat{i} + y\hat{j} + z\hat{k}) = 27\)
\(3x + 3y + 3z = 27\)
Dividing the entire equation by 3, we get our first linear equation:
\(x + y + z = 9\) (Equation 1)
Using the Cross Product Condition (\(\vec{a} \times \vec{b} = 9\vec{c}\))
First, let's calculate the cross product \(\vec{a} \times \vec{b}\). The cross product of two vectors can be calculated using a determinant:
\(\vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & 3 & 3 \\ x & y & z \end{vmatrix}\)
Expanding the determinant:
\(\vec{a} \times \vec{b} = \hat{i}((3)(z) - (3)(y)) - \hat{j}((3)(z) - (3)(x)) + \hat{k}((3)(y) - (3)(x))\)
\(\vec{a} \times \vec{b} = (3z - 3y)\hat{i} + (3x - 3z)\hat{j} + (3y - 3x)\hat{k}\)
Next, calculate \(9\vec{c}\):
\(9\vec{c} = 9(\hat{j} - \hat{k}) = 0\hat{i} + 9\hat{j} - 9\hat{k}\)
Now, we equate the calculated cross product \(\vec{a} \times \vec{b}\) with \(9\vec{c}\) by comparing the coefficients of \(\hat{i}\), \(\hat{j}\), and \(\hat{k}\):
Coefficient of \(\hat{i}\): \(3z - 3y = 0 \implies z - y = 0 \implies z = y\) (Equation 2)
Coefficient of \(\hat{j}\): \(3x - 3z = 9 \implies x - z = 3\) (Equation 3)
Coefficient of \(\hat{k}\): \(3y - 3x = -9 \implies y - x = -3 \implies x - y = 3\) (Equation 4)
Notice that Equation 3 and Equation 4 are consistent if \(y=z\), which is already given by Equation 2.
Solving the System of Equations
We have a system of linear equations:
- \(x + y + z = 9\)
- \(y = z\)
- \(x - y = 3\)
Substitute Equation 2 (\(y = z\)) into Equation 1:
\(x + y + y = 9\)
\(x + 2y = 9\) (Equation 5)
Now we have a simpler system involving only x and y:
- \(x + 2y = 9\) (Equation 5)
- \(x - y = 3\) (Equation 4)
We can solve this system using elimination. Subtract Equation 4 from Equation 5:
\((x + 2y) - (x - y) = 9 - 3\)
\(x + 2y - x + y = 6\)
\(3y = 6\)
\(y = 2\)
Now that we have the value of y, we can find x using Equation 4:
\(x - y = 3\)
\(x - 2 = 3\)
\(x = 5\)
Finally, using Equation 2, we find z:
\(z = y\)
\(z = 2\)
So, the components of vector \(\vec{b}\) are \(x=5\), \(y=2\), and \(z=2\).
Therefore, the vector \(\vec{b}\) is:
\(\vec{b} = 5\hat{i} + 2\hat{j} + 2\hat{k}\)
Verifying the Solution
Let's check if this vector \(\vec{b}\) satisfies the original conditions.
Condition 1: \(\vec{a} \cdot \vec{b} = 27\)
\((3\hat{i} + 3\hat{j} + 3\hat{k}) \cdot (5\hat{i} + 2\hat{j} + 2\hat{k}) = (3)(5) + (3)(2) + (3)(2) = 15 + 6 + 6 = 27\). This condition is satisfied.
Condition 2: \(\vec{a} \times \vec{b} = 9\vec{c}\)
We calculated \(\vec{a} \times \vec{b} = (3z - 3y)\hat{i} + (3x - 3z)\hat{j} + (3y - 3x)\hat{k}\). Substituting \(x=5, y=2, z=2\):
\(\vec{a} \times \vec{b} = (3(2) - 3(2))\hat{i} + (3(5) - 3(2))\hat{j} + (3(2) - 3(5))\hat{k}\)
\(\vec{a} \times \vec{b} = (6 - 6)\hat{i} + (15 - 6)\hat{j} + (6 - 15)\hat{k}\)
\(\vec{a} \times \vec{b} = 0\hat{i} + 9\hat{j} - 9\hat{k}\)
Also, \(9\vec{c} = 9(\hat{j} - \hat{k}) = 9\hat{j} - 9\hat{k}\). This condition is also satisfied.
Both conditions are satisfied by \(\vec{b} = 5\hat{i} + 2\hat{j} + 2\hat{k}\).
Comparing with Options
Let's look at the given options:
- \(3 \hat{i}+4 \hat{j}+2 \hat{k}\)
- \(5 \hat{i}+2 \hat{j}+2 \hat{k}\)
- \(5 \hat{i}-2 \hat{j}+6 \hat{k}\)
- \(3 \hat{i}+3 \hat{j}+4 \hat{k}\)
Our calculated vector \(\vec{b} = 5\hat{i} + 2\hat{j} + 2\hat{k}\) matches option 2.
| Option | Vector \(\vec{b}\) | \(\vec{a} \cdot \vec{b}\) | \(\vec{a} \times \vec{b}\) | Matches Conditions? |
|---|---|---|---|---|
| 1 | \(3 \hat{i}+4 \hat{j}+2 \hat{k}\) | \(3(3)+3(4)+3(2) = 9+12+6 = 27\) (Matches \(\vec{a} \cdot \vec{b}\)) | \(\begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & 3 & 3 \\ 3 & 4 & 2 \end{vmatrix} = \hat{i}(6-12) - \hat{j}(6-9) + \hat{k}(12-9) = -6\hat{i}+3\hat{j}+3\hat{k}\) (Does not match \(9\vec{c}\)) | No |
| 2 | \(5 \hat{i}+2 \hat{j}+2 \hat{k}\) | \(3(5)+3(2)+3(2) = 15+6+6 = 27\) (Matches \(\vec{a} \cdot \vec{b}\)) | \(\begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & 3 & 3 \\ 5 & 2 & 2 \end{vmatrix} = \hat{i}(6-6) - \hat{j}(6-15) + \hat{k}(6-15) = 0\hat{i}+9\hat{j}-9\hat{k}\) (Matches \(9\vec{c}\)) | Yes |
| 3 | \(5 \hat{i}-2 \hat{j}+6 \hat{k}\) | \(3(5)+3(-2)+3(6) = 15-6+18 = 27\) (Matches \(\vec{a} \cdot \vec{b}\)) | \(\begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & 3 & 3 \\ 5 & -2 & 6 \end{vmatrix} = \hat{i}(18-(-6)) - \hat{j}(18-15) + \hat{k}(-6-15) = 24\hat{i}-3\hat{j}-21\hat{k}\) (Does not match \(9\vec{c}\)) | No |
| 4 | \(3 \hat{i}+3 \hat{j}+4 \hat{k}\) | \(3(3)+3(3)+3(4) = 9+9+12 = 30\) (Does not match \(\vec{a} \cdot \vec{b}\)) | - | No |
Conclusion
Based on the calculations using both the dot product and cross product conditions, the vector \(\vec{b}\) is \(5\hat{i} + 2\hat{j} + 2\hat{k}\).
Revision Table: Vector Dot Product and Cross Product
| Concept | Definition | Properties | Geometric Interpretation |
|---|---|---|---|
| Dot Product (\(\vec{a} \cdot \vec{b}\)) | Scalar product: \(\vec{a} \cdot \vec{b} = |\vec{a}||\vec{b}|\cos\theta\) or \(a_x b_x + a_y b_y + a_z b_z\) | Commutative (\(\vec{a} \cdot \vec{b} = \vec{b} \cdot \vec{a}\)), Distributive, \(\vec{a} \cdot \vec{a} = |\vec{a}|^2\) | Measures the extent to which vectors are in the same direction. \(\vec{a} \cdot \vec{b} = 0\) if vectors are orthogonal (perpendicular). |
| Cross Product (\(\vec{a} \times \vec{b}\)) | Vector product: \(\vec{a} \times \vec{b} = |\vec{a}||\vec{b}|\sin\theta \hat{n}\) where \(\hat{n}\) is orthogonal to both \(\vec{a}\) and \(\vec{b}\) (Right-hand rule). Calculated using a determinant of components. | Anti-commutative (\(\vec{a} \times \vec{b} = -\vec{b} \times \vec{a}\)), Distributive, \(\vec{a} \times \vec{a} = \vec{0}\) | Results in a vector perpendicular to the plane formed by \(\vec{a}\) and \(\vec{b}\). Its magnitude equals the area of the parallelogram formed by \(\vec{a}\) and \(\vec{b}\). \(\vec{a} \times \vec{b} = \vec{0}\) if vectors are parallel. |
Additional Information: Applications of Vector Operations
Vector operations like the dot product and cross product are fundamental in physics and engineering. Here are some common applications:
- Work Done (Physics): The work done by a force \(\vec{F}\) over a displacement \(\vec{d}\) is given by the dot product \(W = \vec{F} \cdot \vec{d}\).
- Torque (Physics): The torque \(\vec{\tau}\) produced by a force \(\vec{F}\) applied at a position \(\vec{r}\) from the pivot point is given by the cross product \(\vec{\tau} = \vec{r} \times \vec{F}\).
- Magnetic Force (Physics): The magnetic force \(\vec{F}\) on a charge \(q\) moving with velocity \(\vec{v}\) in a magnetic field \(\vec{B}\) is given by \(\vec{F} = q(\vec{v} \times \vec{B})\).
- Finding Area and Volume (Geometry): The magnitude of the cross product \(|\vec{a} \times \vec{b}|\) gives the area of the parallelogram formed by vectors \(\vec{a}\) and \(\vec{b}\). The scalar triple product \(|\vec{a} \cdot (\vec{b} \times \vec{c})|\) gives the volume of the parallelepiped formed by vectors \(\vec{a}\), \(\vec{b}\), and \(\vec{c}\).
- Determining Vector Relationships: As shown in this problem, dot and cross products can be used to set up equations to find unknown vectors or determine relationships between vectors (e.g., orthogonality, parallelism).
Similar Questions
A vector \(\vec r=a \hat i+b \hat j\) is equally inclined to both x and y axes. If the magnitude of the vector is 2 units, then what are the values of a and b respectively?
Let \(\vec{\text{a}}\) and \(\vec{\text{b}}\) are two unit vectors such that \(\vec{\text{a}}+2 \vec{\text{b}}\) and \(5\vec{\text{a}}−4\vec{\text{b}}\) are perpendicular. What is the angle between \(\vec{\text{a}}\) and \(\vec{\text{b}}\) ?
- The position vectors of vertices A, B and C of triangle ABC are respectively \(\hat{\text{j}}+\hat{\text{k}}, 3\hat{\text{i}}+\hat{\text{j}+5\hat{\text{k}}}\) and \(3\hat{\text{j}}+3\hat{\text{k}}\) . What is angle C equal to?
ABCDEFGH is a cuboid with base ABCD. Let A(0, 0, 0), B(12, 0, 0), C(12, 6, 0) and G(12, 6, 4) be the vertices. If α is the angle between AB and AG; β is the angle between AC and AG, then what is the value of cos 2α + cos 2β?
Consider the following equations for two vectors \(\vec{a}\) and \(\vec{b}\)
1. \(\left( \vec{a}+\vec{b} \right)\cdot \left( \vec{a}-\vec{b} \right)={{\left| {\vec{a}} \right|}^{2}}-{{\left| {\vec{b}} \right|}^{2}}\)
2. \(\left( \left| \vec{a}+\vec{b} \right| \right)\left( \left| \vec{a}-\vec{b} \right| \right)={{\left| {\vec{a}} \right|}^{2}}-{{\left| {\vec{b}} \right|}^{2}}\)
3. \({{\left| \vec{a}\cdot \vec{b} \right|}^{2}}+{{\left| \vec{a}\times \vec{b} \right|}^{2}}={{\left| {\vec{a}} \right|}^{2}}{{\left| {\vec{b}} \right|}^{2}}\)
Which of the above statement are correct?Consider the following statements:
1. The magnitude of \(\vec{a}\times \vec{b}\) is same as the area of a triangle with sides \(\vec{a}\) and \(\vec{b}\)
2. If \(\vec{a}\times \vec{b}=\vec{0}\) where \(\vec{a}\ne \vec{0},~\vec{b}\ne \vec{0},\) then \(\vec{a}=\lambda \vec{b}\)
Which of the above statement is/are correct?If \(\vec{a}\:and\:\vec{b}\) are unit vectors and θ is the angle between them, then what is \({{\sin }^{2}}\left( \frac{\theta }{2} \right)\) equal to?
If in a right-angled triangle ABC, hypotenuse AC = p, then what is \(\overrightarrow {AB} \cdot \overrightarrow {AC} + \overrightarrow {BC} \; \cdot \overrightarrow {BA} + \overrightarrow {CA} \cdot \overrightarrow {CB} \) equal to?
If \(\vec r\) = xî + yĵ + zk̂, then what is \(\vec r\) . (î + ĵ + k̂ ) equal to?
A unit vector perpendicular to each of the vectors 2î - ĵ + k̂ and 3î - 4ĵ - k̂ is
Important Questions from Scalar and Vector Product
If \(\overrightarrow a \) and \(\overrightarrow b\) are two unit vectors inclined to x - axis at angles 30° and 120°, then \(\left| {\overrightarrow a + \overrightarrow b } \right|\) equals
The value of \(\left( {\overrightarrow a - \overrightarrow b } \right).\left[ {\left( {\overrightarrow b - \overrightarrow c } \right) \times \left( {\overrightarrow c - \overrightarrow a } \right)} \right]\) is:
The work done in moving an object along a vector \(\widehat d = 3\widehat i + 2\widehat j - 5\widehat k\) (if the applied force is \(\overrightarrow F = 2\widehat i - \widehat j - \widehat k\)) is
If \(\bar a\) and \(\bar b\) are unit vectors and θ is the angle between them then \(\left| {\frac{{\bar a - \bar b}}{2}} \right|\) is
Two forces F̅1 = î - ĵ + k̂ and F̅2 = 4î + 2ĵ + 3k̂ act on a particle and displace it from the point (0, 1, 2) to (1, -2, 3), then the total work done is
