Question

Three vectors a, b and c are given. Find the equation of a vector that lies in the plane of vector a and vector b and whose projection on vector c is 1/√3.

Answer 1

To find the equation of a vector that lies in the plane of vectors a and b and has a projection of 1/√3 on vector c, we can use vector projection and vector addition.

Let's denote the vector we want to find as \( v \).
The projection of \( v \) on vector \( c \) is given by: \[ \text{proj}_c(v) = \frac{v \cdot c}{||c||} \]

Step 1: Set up the projection equation
Since we want the projection to be \( \frac{1}{\sqrt{3}} \), we can write the equation: \[ \frac{v \cdot c}{||c||} = \frac{1}{\sqrt{3}} \]

Step 2: Express \( v \) as a linear combination of \( a \) and \( b \)
Since \( v \) lies in the plane of vectors \( a \) and \( b \), we can express \( v \) as a linear combination of \( a \) and \( b \): \[ v = \lambda a + \mu b \] where \( \lambda \) and \( \mu \) are scalar coefficients.

Step 3: Substitute \( v \) into the projection equation
Now, substitute \( v = \lambda a + \mu b \) into the projection equation: \[ \frac{(\lambda a + \mu b) \cdot c}{||c||} = \frac{1}{\sqrt{3}} \] Expanding the dot product: \[ \frac{\lambda (a \cdot c) + \mu (b \cdot c)}{||c||} = \frac{1}{\sqrt{3}} \]

Step 4: Use the cross product to find a relation between \( a \), \( b \), and \( v \)
Since \( v \) lies in the plane of \( a \) and \( b \), we can also express \( v \) as: \[ v = (a \times b) \times a \] where \( \times \) denotes the cross product. This means that \( v \) is perpendicular to \( a \) and lies in the plane of \( a \) and \( b \). Therefore, we can take the cross product of \( a \) and \( v \) to obtain \( b \). Hence: \[ v \times a = a \times b \] Solving for \( v \), we get: \[ v = \frac{(a \times b) \times a}{||a||^2} \]

Step 5: Substitute this expression for \( v \) into the projection equation
Now, substitute this expression for \( v \) into the projection equation: \[ \frac{((a \times b) \times a) \cdot c}{||a||^2 ||c||} = \frac{1}{\sqrt{3}} \] Expanding the double cross product: \[ \frac{((b \cdot a) a - (a \cdot a) b) \cdot c}{||a||^2 ||c||} = \frac{1}{\sqrt{3}} \]

Step 6: Simplify the equation
Simplifying the expression: \[ \frac{(b \cdot a) (a \cdot c) - (a \cdot a) (b \cdot c)}{||a||^2 ||c||} = \frac{1}{\sqrt{3}} \]

Step 7: Solve for the scalar coefficients \( \lambda \) and \( \mu \)
This equation provides a relationship between the scalar coefficients \( \lambda \) and \( \mu \). We can now choose any value for \( \mu \) that satisfies the equation and then solve for the corresponding value of \( \lambda \). The vector \( v \) will then be given by: \[ v = \lambda a + \mu b \]