Question

If a→=3i^-2j^-k^ and b→=i^+4j^+k^, find a unit vector n^ normal to the plane containing a→ and b→ such that a→, b→ and n^, in this order, form a right-handed system.

Answer 1

vector a = 3i - 2j - k and vector b = i + 4j + k, the unit vector, n, normal to the plane is (7i - j + 7k)/3√11

What is a vector?

A vector is a quantity having direction as well as magnitude.

Vectors can especially be used in determining the position of one point in space relative to another.

From the question,

The normal vector to the plane is c = a × b

we have that, a = 3i - 2j - k and b = i + 4j + k,

c = a × b

c = (3i - 2j - k) × (i + 4j + k)

= 3i × i + (- 2j) × i + (-k) × i + 3i × 4j + (- 2j) × 4j + (-k) × 4j + 3i × k + (- 2j) × k + (-k) × k

= 3i × i - 2j × i - k × i + 12i × j - 8j × j - 4k × 4j + 3i × k - 2j × k - k × k

It is known that

i × i = 0,

j × j = 0 and

k × k = 0.

we also have that

c = 3(0) - 2(-k) - (-j) + 12k - 8(0) - 16(-i) + 3(-j) - 2i - 0

c = 0 + 2k + j + 12k - 0 + 16i - 3j - 2i

c = 16i - 2i + j - 3j + 2k + 12k

c = 14i -2j + 14k

The unit vector normal to the plane n = c/ |c|

where |c| is the magnitude of c

= √(14² + (-2)² + 14²) = √(196 + 4 + 196)

= √396

= √(36 × 11)

= 6√11

we can write that n = c/|c|

= (14i -2j + 14k)/6√11

we simplify further

= 2(7i - j + 7k)/6√11

= (7i - j + 7k)/3√11

So, the unit vector normal to the plane is (7i - j + 7k)/3√11

Learn more about vectors at: brainly.com/question/25705666

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