Find two vectors that are orthogonal to the vector u, pointing in the opposite directions.&nbsp;U=−14i+32j

Question

Find two vectors that are orthogonal to the vector u, pointing in the opposite directions.&amp;nbsp;U=−14i+32j

Neelam Wainwright · Accepted Answer

If the dot product of two vectors equals 0, then they are orthogonal. Every vector has an endless number of orthogonal counterparts.We are seeking v in such a way as to:u*v=0&amp;nbsp;⟨−14,32⟩⋅⟨v1,v2⟩=0&amp;nbsp;−14v1+32v2=0&amp;nbsp;⇒v1=6v2&amp;nbsp;Now we have that the vector&amp;nbsp;v=⟨6v2,v2⟩&amp;nbsp;is orthogonal to u, where&amp;nbsp;v2∈R&amp;nbsp;is arbitrary.&amp;nbsp;Which means if we take&amp;nbsp;v21=1&amp;nbsp;&amp;nbsp;and&amp;nbsp;&amp;nbsp;v22=−1&amp;nbsp;we have two vectors&amp;nbsp;v1=⟨6,1⟩&amp;nbsp;and&amp;nbsp;v2=⟨−6,−1⟩&amp;nbsp;which are in opposition to one another and orthogonal&amp;nbsp;

Andre BalkonE · Accepted Answer

Let v1 and v2 be the orthogonal vectors we are looking for.To obtain an orthogonal vector, we can take the dot product of u with v1 and set it equal to zero:u·v1=0Since u=−14i+32j, we have:(−14i+32j)·v1=0Simplifying the dot product, we get:−14v1x+32v1y=0Solving this equation, we can express v1 in terms of a free variable t as:v1=32ti+14tjTo find a vector pointing in the opposite direction, we can negate v1:−v1=−32ti−14tjNow, we can choose a specific value for t to obtain two orthogonal vectors pointing in opposite directions. For example, let&#039;s take t=1:v1=32i+14jv2=−32i−14jTherefore, two vectors orthogonal to u=−14i+32j and pointing in the opposite directions are:v1=32i+14jv2=−32i−14j

Jazz Frenia · Accepted Answer

Result:𝐯=𝐢−16𝐣𝐰=−𝐢+16𝐣Solution:Let&#039;s denote the orthogonal vectors as 𝐯 and 𝐰. To ensure they are orthogonal to 𝐮, the dot product of 𝐮 with each of these vectors should be zero. Additionally, we want them to point in opposite directions, which means they will have opposite signs.First, let&#039;s find 𝐯. We can set 𝐯=𝐢−a𝐣, where a is a scalar constant. Taking the dot product of 𝐮 and 𝐯 should yield zero:𝐮·𝐯=(−14𝐢+32𝐣)·(𝐢−a𝐣)=0Expanding the dot product, we have:−14𝐢·𝐢+32𝐣·𝐢−14𝐢·(−a𝐣)+32𝐣·(−a𝐣)=0Using the properties of dot products, we know that 𝐢·𝐢=1, 𝐣·𝐢=0, and 𝐣·𝐣=1. Substituting these values, we get:−14(1)+32(0)+14(a)(0)+32(1)(a)=0Simplifying the equation, we have:−14+32a=0To solve for a, we can multiply through by 4 to eliminate the fraction:−1+6a=0Adding 1 to both sides and then dividing by 6, we find:a=16Therefore, 𝐯=𝐢−16𝐣.Now, let&#039;s find 𝐰. We can set 𝐰=−𝐢+b𝐣, where b is a scalar constant. Taking the dot product of 𝐮 and 𝐰 should yield zero:𝐮·𝐰=(−14𝐢+32𝐣)·(−𝐢+b𝐣)=0Expanding the dot product, we have:−14𝐢·(−𝐢)+32𝐣·(−𝐢)−14𝐢·(b𝐣)+32𝐣·(b𝐣)=0Using the properties of dot products, we find:−14(1)+32(0)−14(b)(0)+32(1)(b)=0Simplifying the equation, we have:−14+32b=0To solve for b, we can multiply through by 4:−1+6b=0Adding 1 to both sides and then dividing by 6, we find:b=16Therefore, 𝐰=−𝐢+16𝐣.

fudzisako · Accepted Answer

First, we need to find a vector that is orthogonal to 𝐮. Let&#039;s call this vector v1. Since 𝐮 is in the form a𝐢+b𝐣, a vector orthogonal to it can be obtained by swapping the coefficients and changing one of the signs. So, v1=b𝐢−a𝐣.Substituting the values of a and b from 𝐮, we have v1=32𝐢+14𝐣.To find the second vector v2, we can take the negative of v1: v2=−v1=−(32𝐢+14𝐣).Simplifying, we have v2=−32𝐢−14𝐣.Therefore, the two vectors orthogonal to 𝐮 and pointing in the opposite directions are v1=32𝐢+14𝐣 and v2=−32𝐢−14𝐣.

Find two vectors in opposite directions that are orthogonal to the vector u.

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