Find parametric equations and symmetric equations for the line through P0 and perpendicular to both given vectors. (P0 corresponds to t = 0.)
P0 = (4, 3, 0)
i + j and j + k
Not exactly sure what to do, how do I find what is perpendicular to both vectors?
First find a vector perpendicular to both P1=i+j(1,1,0) and P2=j+k (0,1,1) by taking the cross product of P1 and P2:
|i j k|
|1 1 0|
|0 1 1|
By evaluating the above determinant which represents the cross product of P1 and P2, we get
P3 = i(1)-j(1)+k(1) = i-j+k =(1,-1,1)
Take any point P(x,y,z) on the plane perpendicular to P3 and passing through P0, the vector P-P0 should be normal to P3, so we have
P-P0 = (x,y,z)-(4,3,0)
P3 = (1,-1,1)
For perpendicularity:
P3.(P3-P0)=0 (dot product)
(1,-1,1).(P-P0) = 0
or
P.(1,-1,1) = P0.(1,-1,1)
using properties of distributivity and commutativity.
(x,y,z).(1,-1,1) = (4,3,0).(1,-1,1)
=>
x-y+z = 1
The required plane is therefore
x -y +z -1 = 0
Checks:
4 -3 +0 -1 =0 => P0 is on the plane
(1,1,0).(1,-1,0)=0 => perp. to i+j
(0,1,1).(1,-1,0)=0 => perp. to j+k
Well, finding a vector that is perpendicular to two given vectors involves taking their cross product. Let's call the given vectors V1 and V2. So V1 = i + j and V2 = j + k.
To find a vector that is perpendicular to both V1 and V2, we can take their cross product:
V_perpendicular = V1 × V2
Calculating the cross product, we have:
V_perpendicular = (i + j) × (j + k)
= (i × j) + (i × k) + (j × j) + (j × k)
Now, we know that i × j = k and j × k = -i (using the right-hand rule). Also, i × i = j × j = 0. So let's substitute these values:
V_perpendicular = k + (i × k) + 0 + (-i)
= k + (-j) + (-i)
= -i - j + k
So we have our perpendicular vector V_perpendicular = -i - j + k.
Now, to find the parametric equations and symmetric equations for the line through P0 and perpendicular to both V1 and V2, we can use the vector form of a line.
Parametric Equations:
x = 4 + (-t)
y = 3 + (-t)
z = 0 + t
Symmetric Equations:
x - 4 = -(y - 3)
x - 4 = -(z - 0)
To find a vector that is perpendicular to both given vectors, you can take their cross product. The cross product of two vectors gives you a vector that is perpendicular to both of them.
Given the two vectors i + j and j + k, we can find their cross product as follows:
(i + j) x (j + k) = (i * k - j * j) - (i * j - j * k) + (i * j - j * k)
= (k - 0) - (0 - j) + (0 - 0)
= k + j
So, the vector that is perpendicular to both i + j and j + k is k + j.
Now, we can use this perpendicular vector to find the parametric and symmetric equations for the line through P0 (4, 3, 0).
Parametric Equations:
Let's take P0 as the starting point on the line and the perpendicular vector k + j as the direction vector of the line. We can write the parametric equations as:
x = 4 + t
y = 3 + t
z = 0 + 0
where t is a parameter that varies along the line.
Symmetric Equations:
The symmetric equations can be obtained by eliminating the parameter t from the parametric equations. In this case, we have:
x - 4 = y - 3 = z - 0
So, the symmetric equations for the line are:
x - 4 = y - 3 = z
I hope this helps! Let me know if you have any further questions.
To find a vector that is perpendicular to both given vectors, you can use the cross product of the two vectors.
Let's start by finding the cross product of the two given vectors:
v1 = i + j
v2 = j + k
To find the cross product, follow these steps:
Step 1: Write the given vectors in their vector form:
v1 = (1, 1, 0)
v2 = (0, 1, 1)
Step 2: Take the cross product of the two vectors using the determinant method:
Cross product = (v1 × v2) = (1 * 1 - 1 * 0) i - (1 * 1 - 0 * 0) j + (1 * 1 - 0 * 1) k
= i - j + k
So we have a vector that is perpendicular to both v1 and v2: (1, -1, 1).
Next, we can find the parametric equations and symmetric equations for the line passing through P0 (4, 3, 0) and perpendicular to both v1 and v2.
Parametric Equations:
Let the line be represented by the point Q(x, y, z).
We can write the position vector P0Q as (x - 4, y - 3, z - 0).
Since the line is perpendicular to both v1 and v2, the dot product of the position vector P0Q and the vector (1, -1, 1) will be zero.
(x - 4, y - 3, z) · (1, -1, 1) = 0
Expanding this equation, we get:
(x - 4) + (-1)(y - 3) + z = 0
Simplifying further, we have:
x - 4 - y + 3 + z = 0
x - y + z - 1 = 0
These are the symmetric equations for the line.
To obtain the parametric equations, let's set z = t (a parameter).
Then, we have:
x - y + t - 1 = 0
Rearranging the terms, we get:
x = y - t + 1
So, the parametric equations for the line passing through P0 and perpendicular to both v1 and v2 are:
x = y - t + 1
z = t
By substituting t with any real number, you can generate points on the line.