A unit vector perpendicular to each of the vectors 2î - ĵ + k̂ and 3î - 4ĵ - k̂ is
Finding a Unit Vector Perpendicular to Two Vectors
To find a unit vector that is perpendicular to two given vectors, say vector A and vector B, we can use the cross product. The cross product of two vectors, A × B, results in a vector that is perpendicular to both A and B. Once we have this perpendicular vector, we can find its unit vector by dividing the vector by its magnitude.
Given Vectors
We are given two vectors:
- Vector A = \(2\hat{i} - \hat{j} + \hat{k}\)
- Vector B = \(3\hat{i} - 4\hat{j} - \hat{k}\)
Calculating the Cross Product
The cross product A × B is calculated using the determinant of a matrix:
\[ \vec{A} \times \vec{B} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & -1 & 1 \\ 3 & -4 & -1 \end{vmatrix} \]Expanding the determinant:
\[ \vec{A} \times \vec{B} = \hat{i}((-1)(-1) - (1)(-4)) - \hat{j}((2)(-1) - (1)(3)) + \hat{k}((2)(-4) - (-1)(3)) \] \[ \vec{A} \times \vec{B} = \hat{i}(1 + 4) - \hat{j}(-2 - 3) + \hat{k}(-8 + 3) \] \[ \vec{A} \times \vec{B} = \hat{i}(5) - \hat{j}(-5) + \hat{k}(-5) \] \[ \vec{A} \times \vec{B} = 5\hat{i} + 5\hat{j} - 5\hat{k} \]Let the resulting perpendicular vector be C = \(5\hat{i} + 5\hat{j} - 5\hat{k}\).
| Component | Calculation | Result |
|---|---|---|
| \(\hat{i}\) | \((-1)(-1) - (1)(-4)\) | \(1 + 4 = 5\) |
| \(\hat{j}\) | \(-\left( (2)(-1) - (1)(3) \right)\) | \(-\left( -2 - 3 \right) = -(-5) = 5\) |
| \(\hat{k}\) | \((2)(-4) - (-1)(3)\) | \(-8 + 3 = -5\) |
Finding the Magnitude of the Perpendicular Vector
The magnitude of vector C = \(5\hat{i} + 5\hat{j} - 5\hat{k}\) is calculated as:
\[ |\vec{C}| = \sqrt{(5)^2 + (5)^2 + (-5)^2} \] \[ |\vec{C}| = \sqrt{25 + 25 + 25} \] \[ |\vec{C}| = \sqrt{75} \] \[ |\vec{C}| = \sqrt{25 \times 3} \] \[ |\vec{C}| = 5\sqrt{3} \]Finding the Unit Vector
A unit vector in the direction of C is given by \(\hat{C} = \frac{\vec{C}}{|\vec{C}|}\):
\[ \hat{C} = \frac{5\hat{i} + 5\hat{j} - 5\hat{k}}{5\sqrt{3}} \] \[ \hat{C} = \frac{5(\hat{i} + \hat{j} - \hat{k})}{5\sqrt{3}} \] \[ \hat{C} = \frac{1}{\sqrt{3}}(\hat{i} + \hat{j} - \hat{k}) \] \[ \hat{C} = \frac{1}{\sqrt{3}}\hat{i} + \frac{1}{\sqrt{3}}\hat{j} - \frac{1}{\sqrt{3}}\hat{k} \]This unit vector is perpendicular to both of the original vectors.
Comparing with Options
Let's compare our calculated unit vector with the given options:
- Option 1: \(\frac{1}{{\sqrt 3 }}\hat i + \frac{1}{{\sqrt 3 }}\hat j - \frac{1}{{\sqrt 3 }}\hat k\)
- Option 2: \(\frac{1}{{\sqrt 2 }}\hat i + \frac{1}{2}\hat j + \frac{1}{2}\hat k\)
- Option 3: \(\frac{1}{{\sqrt 3 }}\hat i - \frac{1}{{\sqrt 3 }}\hat j - \frac{1}{{\sqrt 3 }}\hat k\)
- Option 4: \(\frac{1}{{\sqrt 3 }}\hat i - \frac{1}{{\sqrt 3 }}\hat j + \frac{1}{{\sqrt 3 }}\hat k\)
Our calculated unit vector matches Option 1.
Conclusion
The unit vector perpendicular to \(2\hat{i} - \hat{j} + \hat{k}\) and \(3\hat{i} - 4\hat{j} - \hat{k}\) is \(\frac{1}{{\sqrt 3 }}\hat i + \frac{1}{{\sqrt 3 }}\hat j - \frac{1}{{\sqrt 3 }}\hat k\).
Revision Table: Vector Operations
| Operation | Description | Use Case |
|---|---|---|
| Dot Product (\(\vec{A} \cdot \vec{B}\)) | Scalar result; measures how much one vector extends in the direction of another. \(\vec{A} \cdot \vec{B} = |\vec{A}||\vec{B}|\cos\theta\). If \(\vec{A} \cdot \vec{B} = 0\) (and \(\vec{A}, \vec{B}\) are non-zero), vectors are perpendicular. | Finding angle between vectors, checking for orthogonality, calculating work done. |
| Cross Product (\(\vec{A} \times \vec{B}\)) | Vector result; perpendicular to both vectors. Magnitude is \(|\vec{A}||\vec{B}|\sin\theta\). Direction follows right-hand rule. If \(\vec{A} \times \vec{B} = \vec{0}\) (and \(\vec{A}, \vec{B}\) are non-zero), vectors are parallel. | Finding a vector perpendicular to a plane, calculating torque, calculating area of a parallelogram. |
| Unit Vector (\(\hat{A}\)) | A vector with magnitude 1, having the same direction as the original vector. \(\hat{A} = \frac{\vec{A}}{|\vec{A}|}\). | Representing direction, normalizing vectors. |
Additional Information: Properties of Unit Vectors and Perpendicularity
A unit vector is a fundamental concept in vector algebra. It simplifies calculations when only the direction of a vector is important. The process of finding a unit vector involves normalizing a vector by dividing it by its magnitude. This preserves the direction but scales the magnitude to exactly 1.
Two vectors are perpendicular or orthogonal if the angle between them is 90 degrees. In terms of vector operations:
- The dot product of two non-zero perpendicular vectors is zero (\(\vec{A} \cdot \vec{B} = 0\)).
- The cross product of two non-zero perpendicular vectors has a magnitude equal to the product of their individual magnitudes (\(|\vec{A} \times \vec{B}| = |\vec{A}||\vec{B}|\)). The resulting vector is perpendicular to the plane containing the two vectors.
The cross product is particularly useful for finding vectors that are mutually perpendicular to two given vectors, as demonstrated in this problem. The direction of the cross product \(\vec{A} \times \vec{B}\) is given by the right-hand rule. Note that \(\vec{B} \times \vec{A} = -(\vec{A} \times \vec{B})\), meaning it points in the opposite direction, but is still perpendicular to both \(\vec{A}\) and \(\vec{B}\). Therefore, both \(\hat{C}\) and \(-\hat{C}\) are unit vectors perpendicular to the original two vectors. In this case, only one of these possibilities was provided in the options.
Similar Questions
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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
