if theta is an acute angle between the vectors i+j+k and 2i-j+k then tan theta is
Yes
To find the value of tan(θ), we need to first find the dot product and magnitude of the given vectors.
Let's label the first vector as A = i + j + k and the second vector as B = 2i - j + k.
1. Dot Product:
The dot product of two vectors A and B is given by A • B = |A||B|cos(θ), where θ is the angle between the vectors.
Using the dot product formula, we can find the dot product of the given vectors as follows:
A • B = (1)(2) + (1)(-1) + (1)(1) = 2 - 1 + 1 = 2.
2. Magnitude:
The magnitude (or length) of a vector A is given by |A| = √(A • A).
Calculating the magnitudes of the given vectors:
|A| = √(1^2 + 1^2 + 1^2) = √3.
|B| = √(2^2 + (-1)^2 + 1^2) = √6.
Now, let's find the value of tan(θ).
tan(θ) = |A|sin(θ) / |A|cos(θ) = sin(θ) / cos(θ).
Using the dot product and magnitude values previously found, we have:
tan(θ) = (A • B) / (|A||B|)
= 2 / (√3 √6)
= 2 / (√18)
= 2 / (3√2)
= (2√2) / (3√2)
= √2 / 3.
Therefore, the value of tan(θ) is √2 / 3.
To find the value of tan(theta) between two vectors, you can use the dot product formula and the magnitude of the vectors.
Step 1: Calculate the dot product of the two vectors.
The dot product of two vectors A and B is given by:
A ⋅ B = |A| |B| cos(theta)
where:
A ⋅ B represents the dot product of A and B,
|A| represents the magnitude (or length) of vector A,
|B| represents the magnitude (or length) of vector B, and
cos(theta) represents the cosine of the angle theta between the two vectors.
In this case, we have two vectors:
A = i + j + k and
B = 2i - j + k
The dot product of A and B can be calculated as follows:
A ⋅ B = (1)(2) + (1)(-1) + (1)(1)
= 2 - 1 + 1
= 2
Step 2: Calculate the magnitude of the vectors.
The magnitude (or length) of a vector is calculated using the Pythagorean theorem. For vector V = (x, y, z), the magnitude |V| is given by:
|V| = sqrt(x^2 + y^2 + z^2)
In this case, the magnitude of vector A is:
|A| = sqrt(1^2 + 1^2 + 1^2)
= sqrt(3)
The magnitude of vector B is:
|B| = sqrt(2^2 + (-1)^2 + 1^2)
= sqrt(6)
Step 3: Calculate the cosine of theta.
Using the dot product formula mentioned earlier, we can rearrange the formula to solve for cos(theta):
cos(theta) = (A ⋅ B) / (|A| |B|)
In this case, cos(theta) = 2 / (sqrt(3) * sqrt(6))
Step 4: Calculate tan(theta).
We know that tan(theta) = sin(theta) / cos(theta)
Since theta is an acute angle, we can use the identity sin(theta) = sqrt(1 - cos^2(theta)). Therefore:
tan(theta) = sin(theta) / cos(theta)
= sqrt(1 - cos^2(theta)) / cos(theta)
= sqrt(1 - (2 / (sqrt(3) * sqrt(6)))^2) / (2 / (sqrt(3) * sqrt(6)))
By substituting the values, you can calculate tan(theta).
Well, theta might be an acute angle, but let's not give it an inflated ego by calling it "tan theta." I mean, it's important to stay humble, right? So, instead of "tan theta," let's say "tan(theta) is..." drumroll, please... the magnitude of the cross product of the two vectors divided by the dot product of the two vectors. See? Who needs "tan theta" when you have such a fancy formula?
√3 * √6 cosθ = 1*2 - 1*1 + 1*1
cosθ = 2/√18
so tanθ = √14/2