Question

Determine whether each pair of vectors is orthogonal.i + 3√2j, 6i - √2j

Accepted Answer

Recall that two vectors are orthogonal if their dot product is zero. The dot product of vectors $$ \mathbf{a} = a_1 \mathbf{i} + a_2 \mathbf{j} $$ and $$ \mathbf{b} = b_1 \mathbf{i} + b_2 \mathbf{j}  is $$given by the formula:

$$ 
\mathbf{a} \cdot \mathbf{b} = a_1 b_1 + a_2 b_2 
$$ Identify the components of the given vectors. For the first vector $$ \mathbf{v_1} = \mathbf{i} + 3\sqrt{2} \mathbf{j} $$, the components are $$ a_1 = 1 $$ and $$ a_2 = 3\sqrt{2} . $$For the second vector $$ \mathbf{v_2} = 6 \mathbf{i} - \sqrt{2} \mathbf{j} $$, the components are $$ b_1 = 6 $$ and $$ b_2 = -\sqrt{2} . $$Calculate the dot product using the components:

$$ 
\mathbf{v_1} \cdot \mathbf{v_2} = (1)(6) + (3\sqrt{2})(-\sqrt{2}) 
$$ Simplify the expression by multiplying the terms:

$$ 
(1)(6) = 6 
$$

and

$$ 
(3\sqrt{2})(-\sqrt{2}) = 3 	imes (-1) 	imes (\sqrt{2} 	imes \sqrt{2}) = 3 	imes (-1) 	imes 2 = -6 
$$ Add the results from the two products to find the dot product:

$$ 
6 + (-6) = 0 
$$  
Since the dot product is zero, conclude that the vectors are orthogonal.

Determine whether each pair of vectors is orthogonal.
i + 3√2j, 6i - √2j

비슷한 문제에 대한 검증된 영상 답변:

주요 개념

Vector Components and Representation

Dot Product of Vectors

Orthogonality of Vectors