Determine a vector equation for the line that is perpendicular to the vectors 𝑢 = (2,0,1) and 𝑣 = (0,3, −1) and passes through the point (5,2,1).
You want a vector <a,b,c> which is perpendicular to the two given vectors, that is,
you want the cross-product of the two given vectors.
once you have that , the vector equation would be
r = (5,2,1) + t(a,b,c)
I assume you know how to find the cross-product
let me know what you get
I got (-3, 2, 6) as my cross product.
So the final answer would be r = (5,2,1) + t(-3,2,6).
Thank you so much for your help!
correct
To determine a vector equation for the line that is perpendicular to the vectors 𝑢 = (2,0,1) and 𝑣 = (0,3, −1) and passes through the point (5,2,1), we can follow these steps:
Step 1: Find the cross product of the given vectors 𝑢 and 𝑣. The cross product of two vectors is a vector that is perpendicular to both of these vectors.
The cross product 𝑢 × 𝑣 can be found using the formula:
𝑢 × 𝑣 = (𝑢𝑦𝑣𝑧 − 𝑢𝑧𝑣𝑦, 𝑢𝑧𝑣𝑥 − 𝑢𝑥𝑣𝑧, 𝑢𝑥𝑣𝑦 − 𝑢𝑦𝑣𝑥)
Substituting the values 𝑢 = (2,0,1) and 𝑣 = (0,3, −1) into the formula, we get:
𝑢 × 𝑣 = ((0 × −1) − (1 × 3), (1 × 0) − (2 × −1), (2 × 3) − (0 × 0))
= (-3, 2, 6)
So, the cross product of the vectors 𝑢 and 𝑣 is 𝑢 × 𝑣 = (-3, 2, 6).
Step 2: Use the point (5,2,1) and the direction vector 𝑢 × 𝑣 from Step 1 to find the vector equation for the line.
Let 𝑟 = (𝑥, 𝑦, 𝑧) be a generic point on the line.
Since the line is perpendicular to the vector 𝑢 × 𝑣, we know that the dot product of the direction vector and any vector parallel to the line is zero.
The direction vector of the line is 𝑢 × 𝑣 = (-3, 2, 6). Therefore, the following equation holds:
(-3)(𝑥 − 5) + (2)(𝑦 − 2) + (6)(𝑧 − 1) = 0
Simplifying further, we get:
-3𝑥 + 15 + 2𝑦 - 4 + 6𝑧 - 6 = 0
-3𝑥 + 2𝑦 + 6𝑧 + 5 = 0
Hence, the vector equation for the line perpendicular to the vectors 𝑢 = (2,0,1) and 𝑣 = (0,3, −1) and passing through the point (5,2,1) is:
𝑟 = (𝑥, 𝑦, 𝑧) = (5,2,1) + 𝑡(-3, 2, 6), where 𝑡 is a parameter.