Math Problem Statement

1 Find the value of 𝑐 for which the vectors (1, 𝑐, 1) and (1, √2, 1) are orthogonal (perpendicular).

Solution

To find the value of cc for which the vectors v1=(1,c,1)\mathbf{v_1} = (1, c, 1) and v2=(1,2,1)\mathbf{v_2} = (1, \sqrt{2}, 1) are orthogonal, we need to use the fact that two vectors are orthogonal if their dot product is zero.

The dot product of two vectors v1=(v1x,v1y,v1z)\mathbf{v_1} = (v_{1x}, v_{1y}, v_{1z}) and v2=(v2x,v2y,v2z)\mathbf{v_2} = (v_{2x}, v_{2y}, v_{2z}) is given by:

v1β‹…v2=v1xβ‹…v2x+v1yβ‹…v2y+v1zβ‹…v2z\mathbf{v_1} \cdot \mathbf{v_2} = v_{1x} \cdot v_{2x} + v_{1y} \cdot v_{2y} + v_{1z} \cdot v_{2z}

Substituting the given vectors:

v1β‹…v2=1β‹…1+cβ‹…2+1β‹…1\mathbf{v_1} \cdot \mathbf{v_2} = 1 \cdot 1 + c \cdot \sqrt{2} + 1 \cdot 1

Simplifying:

1+c2+1=2+c21 + c\sqrt{2} + 1 = 2 + c\sqrt{2}

For the vectors to be orthogonal, their dot product must be zero:

2+c2=02 + c\sqrt{2} = 0

Solve for cc:

c2=βˆ’2c\sqrt{2} = -2

c=βˆ’22c = \frac{-2}{\sqrt{2}}

Simplify further:

c=βˆ’22=βˆ’2c = \frac{-2}{\sqrt{2}} = -\sqrt{2}

So, the value of cc is βˆ’2\boxed{-\sqrt{2}}.

Would you like more details or have any questions?


Here are 8 related questions you might find interesting:

  1. What are the general conditions for two vectors to be parallel?
  2. How do you find the angle between two vectors using their dot product?
  3. Can two non-zero vectors be orthogonal and parallel at the same time?
  4. How do you normalize a vector?
  5. What is the geometric interpretation of the cross product of two vectors?
  6. How can you determine if three vectors are coplanar?
  7. What is the significance of the magnitude of a vector in physics?
  8. How do you find the projection of one vector onto another?

Tip: Remember that the dot product of two vectors is a scalar quantity, while the cross product results in a vector.

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Math Problem Analysis

Mathematical Concepts

Vectors
Dot product
Orthogonality

Formulas

Dot product formula

Theorems

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Suitable Grade Level

Grades 10-12