What is the unit vector that is orthogonal to the plane containing $(- i + j + k)$ $(-i + j + k)$ and $(3 i + 2 j - 3 k)$ $(3i + 2j - 3k)$ ?

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What is the unit vector that is orthogonal to the plane containing $(- i + j + k)$ $(-i + j + k)$ and $(3 i + 2 j - 3 k)$ $(3i + 2j - 3k)$ ?

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Answer 1 · 2018-03-13T16:02:01

There are two unit vectors here, depending on your order of operations. They are $(- 5 i + 0 j - 5 k)$ $(-5i +0j -5k)$ and $(5 i + 0 j 5 k)$ $(5i +0j 5k)$

Explanation:

When you take the cross product of two vectors, you are calculating the vector that is orthogonal to the first two. However, the solution of $\to A \otimes \to B$ $vecAoxvecB$ is usually equal and opposite in magnitude of $\to B \otimes \to A$ $vecBoxvecA$ .

As a quick refresher, a cross-product of $\to A \otimes \to B$ $vecAoxvecB$ builds a 3x3 matrix that looks like:

$| i j k |$ $|i j k|$
$∣ ∣ A_{x} A_{y} A_{z} ∣ ∣$ $|A_x A_y A_z|$
$∣ ∣ B_{x} B_{y} B_{z} ∣ ∣$ $|B_x B_y B_z|$

and you get each term by taking the product of the diagonal terms going from left to right, starting from a given unit vector letter (i, j, or k) and subtracting the product of diagonal terms going from right to left, starting from the same unit vector letter:

$(A_{y} \times B_{z} - A_{z} \times B_{y}) i + (A_{z} \times B_{x} - A_{x} \times B z) j + (A_{x} \times B_{y} - A_{y} \times B_{x}) k$ $(A_yxxB_z-A_zxxB_y)i+(A_zxxB_x-A_x xxBz)j+(A_x xxB_y-A_yxxB_x)k$

For the two solutions, lets set:
$\to A = [- i + j + k]$ $vecA=[-i+j+k]$
$\to B = [3 i + 2 j - 3 k]$ $vecB=[3i+2j-3k]$

Let's look at both solutions:

$\to A \otimes \to B$ $vecAoxvecB$

As stated above:

$\to A \otimes \to B = (A_{y} \times B_{z} - A_{z} \times B_{y}) i + (A_{z} \times B_{x} - A_{x} \times B z) j + (A_{x} \times B_{y} - A_{y} \times B_{x}) k$ $vecAoxvecB=(A_yxxB_z-A_zxxB_y)i+(A_zxxB_x-A_x xxBz)j+(A_x xxB_y-A_yxxB_x)k$

$\to A \otimes \to B = (1 \times (- 3) - 1 \times 2) i + (1 \times 3 - (- 1) \times (- 3)) j + (- 1 \times 2 - 1 \times 3) k$ $vecAoxvecB=(1xx(-3)-1xx2)i+(1xx3-(-1)xx(-3))j+(-1 xx2-1xx3)k$

$\to A \otimes \to B = (- 3 - 2) i + (3 - 3) j + (- 2 - 3) k$ $vecAoxvecB=(-3-2)i+(3-3)j+(-2-3)k$

$\to A \otimes \to B = - 5 i + 0 j - 5 k$ $color(red)(vecAoxvecB=-5i+0j-5k$

$\to B \otimes \to A$ $vecBoxvecA$

As a flip to the first formulation, take the diagonals again, but the matrix is formed differently:

$| i j k |$ $|i j k|$
$∣ ∣ B_{x} B_{y} B_{z} ∣ ∣$ $|B_x B_y B_z|$
$∣ ∣ A_{x} A_{y} A_{z} ∣ ∣$ $|A_x A_y A_z|$

$\to B \otimes \to A = (A_{z} \times B_{y} - A_{y} \times B_{z}) i + (A_{x} \times B_{z} - A_{z} \times B x) j + (A_{y} \times B_{x} - A_{x} \times B_{y}) k$ $vecBoxvecA=(A_zxxB_y-A_yxxB_z)i+(A_x xxB_z-A_z xxBx)j+(A_y xxB_x-A_x xxB_y)k$

Notice that the subtractions are flipped around. This is what causes the 'Equal and opposite' form.

$\to B \otimes \to A = (1 \times 2 - 1 \times (- 3)) i + ((- 1) \times (- 3) - 1 \times 3) j + (1 \times 3 - (- 1) \times 2) k$ $vecBoxvecA=(1xx2-1xx(-3))i+((-1) xx(-3)-1 xx3)j+(1 xx3-(-1) xx2)k$

$\to B \otimes \to A = (2 - (- 3)) i + (3 - 3) j + (3 - (- 2)) k$ $vecBoxvecA=(2-(-3))i+(3-3)j+(3-(-2))k$
$\to B \otimes \to A = 5 i + 0 j + 5 k$ $color(blue)(vecBoxvecA=5i+0j+5k$

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What is the unit vector that is orthogonal to the plane containing $(- i + j + k)$ $(-i + j + k)$ and $(3 i + 2 j - 3 k)$ $(3i + 2j - 3k)$ ?

1 Answer

Explanation:

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Science

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What is the unit vector that is orthogonal to the plane containing (-i + j + k) (−i+j+k) (-i + j + k) and (3i + 2j - 3k) (3i+2j−3k) (3i + 2j - 3k) ?

1 Answer

Explanation:

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What is the unit vector that is orthogonal to the plane containing $(- i + j + k)$ $(-i + j + k)$ and $(3 i + 2 j - 3 k)$ $(3i + 2j - 3k)$ ?