Let vector A=i^A cos theta +j^A sin theta be any vector. Another vector vector B which is normal to A is
To find a vector that is normal to vector A, we need to understand what it means for two vectors to be normal or perpendicular to each other.
Two vectors are considered perpendicular or normal to each other if their dot product is zero. The dot product of two vectors A and B can be calculated using the formula:
A · B = |A| |B| cos θ
where |A| and |B| are the magnitudes of vectors A and B, respectively, and θ is the angle between them.
In this case, vector B should be perpendicular to vector A. This means their dot product should be zero:
A · B = 0
Now, let's find vector B using this information.
Given vector A = i^A cos θ + j^A sin θ, we can represent vector B as B = i^B + j^B.
To find B, we need to calculate its components i^B and j^B. We can use the dot product equation to solve for these components.
A · B = 0
(i^A cos θ + j^A sin θ) · (i^B + j^B) = 0
Using the dot product formula, expand the expression:
(i^A cos θ + j^A sin θ) · (i^B + j^B) = i^A · i^B + j^A · j^B + i^A · j^B + j^A · i^B
Since i^A and j^A are perpendicular to each other (i.e. i^A · j^A = 0), and i^B and j^B are also perpendicular to each other (i.e. i^B · j^B = 0), the above expression simplifies to:
0 + 0 + i^A · j^B + 0 = 0
Simplifying further, we get:
i^A · j^B = 0
This means that the dot product of the i-component of A and the j-component of B should be zero.
To satisfy this condition, set i^A · j^B to zero. Since i^A = 1 and j^B = 1, we have:
1 · 1 = 0
This is not possible, as the product of two non-zero numbers cannot be zero.
Hence, vector B cannot be found based on the given conditions.
To find a vector that is normal to vector A, we can use the properties of dot product between two vectors. The dot product of two vectors is zero if the vectors are orthogonal (perpendicular) to each other.
Given vector A = i^A cos(theta) + j^A sin(theta), we can take the dot product of vector A with an unknown vector B = x^B + y^B. The dot product between A and B can be written as:
A • B = (i^A cos(theta) + j^A sin(theta)) • (x^B + y^B)
Expanding the dot product, we have:
A • B = (i^A cos(theta)) • (x^B + y^B) + (j^A sin(theta)) • (x^B + y^B)
Using the properties of dot product, the cross product between unit vectors i and j is zero, so:
A • B = (i^A cos(theta)) • (x^B) + (i^A cos(theta)) • (y^B) + (j^A sin(theta)) • (x^B) + (j^A sin(theta)) • (y^B)
Since we know that i^A and j^A are perpendicular to each other, their dot product is zero:
A • B = 0 + 0 + (j^A sin(theta)) • (x^B) + (j^A sin(theta)) • (y^B)
To make the dot product zero, we need the two terms involving j^A to be zero. This means that:
(j^A sin(theta)) • (x^B) = 0
(j^A sin(theta)) • (y^B) = 0
Since sin(theta) is not equal to zero (because it is a function of theta and can take different values), we can set the coefficients of x^B and y^B to be zero:
(j^A sin(theta)) • (x^B) = 0 ==> x^B = 0
(j^A sin(theta)) • (y^B) = 0 ==> y^B = 0
So, vector B must be of the form B = 0^B + 0^B, which means vector B is the zero vector.
You need A•B = 0
so how about -Asinθ i + Acosθ j