Find the cosine of the angle between the planes x + y + z = 0 and x + 2y + 3z = 1.

Question

Daphne Broadhurst · Accepted Answer

Therefore,
the normal vector to x + y + z=0 is

⟨ 1, 1, 1 ⟩

the normal vector to x + 2y + 3z=1 is

⟨ 1, 2, 3 ⟩

If the angle between the two normals/planes is

θ

, we can write

⟨ 1, 1, 1 ⟩ \cdot ⟨ 1, 2, 3 ⟩ = | ⟨ 1, 1, 1 ⟩ | | ⟨ 1, 2, 3 ⟩ | \cos θ

(1) (1) + (1) (2) + (1) (3) = \sqrt{1^{2} + 1^{2} + 1^{2}} \sqrt{1^{2} + 2^{2} + 3^{2}} \cos θ

1 + 2 + 3 = \sqrt{1 + 1 + 1} \sqrt{1 + 4 + 9} \cos θ

6 = \sqrt{3} \sqrt{14} \cos θ

\cos θ = \frac{6}{\sqrt{3 \cdot 14}} = \frac{6}{\sqrt{42}}

Result:

\cos θ = \frac{6}{\sqrt{42}}

karton · Accepted Answer

Answer:21433Explanation:Let&#039;s consider the first plane with equation x+y+z=0. To find its normal vector, we take the coefficients of x, y, and z. Therefore, the normal vector 𝐚 of the first plane is given by:𝐚=[111]Now, let&#039;s consider the second plane with equation x+2y+3z=1. Its normal vector 𝐛 is obtained by taking the coefficients of x, y, and z:𝐛=[123]The dot product between two vectors is defined as the product of their corresponding components summed up. In this case, we can calculate the dot product between 𝐚 and 𝐛 as follows:𝐚·𝐛=(1)(1)+(1)(2)+(1)(3)=1+2+3=6To find the magnitude (length) of a vector, we use the formula:|𝐯|=∑i=1nvi2Applying this formula to both vectors, we get:|𝐚|=12+12+12=3|𝐛|=12+22+32=14Finally, the cosine of the angle between the planes can be calculated using the dot product and magnitudes:cos(θ)=𝐚·𝐛|𝐚|·|𝐛|=63·14Simplifying the expression, we have:cos(θ)=21433Therefore, the cosine of the angle between the planes x+y+z=0 and x+2y+3z=1 is 21433.

star233 · Accepted Answer

Step 1:First, we need to determine the normal vectors of the planes. The normal vector of a plane with the equation ax+by+cz=d is (a,b,c).For the plane x+y+z=0, the normal vector is (1,1,1).For the plane x+2y+3z=1, the normal vector is (1,2,3).To find the cosine of the angle between the two planes, we normalize the normal vectors, calculate their dot product, and divide it by the product of their magnitudes.Step 2:Normalization of a vector (a,b,c) is done by dividing each component by the magnitude of the vector, a2+b2+c2.So, the cosine of the angle between the two planes is:cos(θ)=n1·n2|n1|·|n2|Substituting the values, we have:cos(θ)=(1,1,1)·(1,2,3)12+12+12·12+22+32Simplifying further, we get:cos(θ)=63·14Therefore, the cosine of the angle between the planes x+y+z=0 and x+2y+3z=1 is 63·14.

alenahelenash · Accepted Answer

To find the cosine of the angle between the planes, we can use the normal vectors of the planes. The normal vector of a plane is the vector perpendicular to the plane's surface.
First, let's find the normal vectors of the given planes. The general form of a plane equation is

A x + B y + C z = D

, where

(A, B, C)

is the normal vector of the plane.
For the plane

x + y + z = 0

, the normal vector is

(1, 1, 1)

.
For the plane

x + 2 y + 3 z = 1

, the normal vector is

(1, 2, 3)

.
Next, we can find the cosine of the angle between the two normal vectors using the dot product. The dot product of two vectors

𝐮

and

𝐯

is defined as

𝐮 \cdot 𝐯 = | 𝐮 | \cdot | 𝐯 | \cdot \cos θ

, where

θ

is the angle between the two vectors.
Let

θ

be the angle between the normal vectors

(1, 1, 1)

and

(1, 2, 3)

. We can find the dot product of these vectors as follows:

𝐮 \cdot 𝐯 = (1, 1, 1) \cdot (1, 2, 3) = 1 \cdot 1 + 1 \cdot 2 + 1 \cdot 3 = 6

.
Next, we need to find the magnitudes (lengths) of the two normal vectors:

| 𝐮 | = \sqrt{1^{2} + 1^{2} + 1^{2}} = \sqrt{3}

.

| 𝐯 | = \sqrt{1^{2} + 2^{2} + 3^{2}} = \sqrt{14}

.
Now we can substitute the dot product and the magnitudes into the dot product formula to find the cosine of the angle:

\cos θ = \frac{𝐮 \cdot 𝐯}{| 𝐮 | \cdot | 𝐯 |} = \frac{6}{\sqrt{3} \cdot \sqrt{14}} = \frac{6}{\sqrt{42}}

.
Therefore, the cosine of the angle between the planes

x + y + z = 0

and

x + 2 y + 3 z = 1

is

\frac{6}{\sqrt{42}}

.

Find the cosine of the angle between the planes x + y + z = 0 and x + 2y + 3z =

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