Find two unit vectors orthogonal to both <1,-1,1> and <0,4,4>.?
Given two vectors u and v, their cross product w = u × v is orthogonal to them both. Then ± w / |w| are two unit vectors orthogonal to both u and v. Here u = 〈1, -1, 1〉 and v = 〈0, 4, 4〉.
Find two unit vectors orthogonal to a=⟨−1,2,1⟩ and b=⟨−4,0,−5⟩?
I got -10i - 9j + 8k; the next step is to make it a unit vector with the same direction. Since |<-10,-9,8>| = sqrt((-10)^2 + (-9)^2 + 8^2) = sqrt(245), take [1/sqrt(245)]<-10,-9,8> for the unit vector. For a second, take [-1/sqrt(245)]<-10,-9,8>.
Find two unit vectors orthogonal to A=(1, 3, 0) B =(2, 0, 5) first vector must have positive first coordinate?
First let's find two vectors. Then we will make them unit vectors by dividing them by their magnitude. Take the cross product and you will get a vector orthogonal to both vector A and vector B. N = A X B = <1, 3, 0> X <2, 0, 5> = <15, -5, -6> Calculate the magnitude of N. || N || = √[15² + (-5)² + (-6)²] = √(225 + 25 + 36) = √286 The first unit vector is: n1 = N / || N || = <15/√286, -5/√286, -6/√286> The other unit vector will be pointing in the opposite direction. Reverse the signs. n2 = -n1 = <-15/√286, 5/√286, 6/√286>
How do you find a unit vector orthogonal to both (1,2,2) and (3,4,5)?
If its orthogonal to both, you have to use the vector cross product, to find the required vector. (1st step of my answer)Then as it has to be a unit vector, i.e. a vector with a modulus of 1, you first find the modulus of the vector you've just obtained and "trim" your vector's modulus down to 1 by simply multiplying it by (1/x) where x was the original modulus.
Find the two unit vectors orthogonal to both a=<3,1,1>and b=<-1,2,1>?
The cross product a x b represents a vector which is perpendicular to both vectors a and b with a direction given by the right-hand rule and a magnitude equal to the area of the parallelogram that the vectors span. But for us the main thing is that c = a x b is perpendicular to a and to b. ...........| i....j... k..| a x b = | 3...1...1..| ...........|-1...2...1..| c = a x b = i (1 - 2) - j (3 + 1 ) + k (6 + 1 ) = -1i - 4 j + 7 k So two vectors c = ( -1; -4; 7) and opposite -c = (1; 4; -7) are perpendicular to the vectors a and b. The last thing - we just need to make them unit vectors. The length |c| = √(1² + 4² + 7² ) = √(1 + 16 + 49) = √66 The unit vectors: ( -1/√66; -4/√66; 7/√66) and ( 1/√66; 4/√66; - 7/√66)
What are the two unit vectors that are orthogonal to (-3, 4)?
What are the two unit vectors that are orthogonal to [math](-3,4).[/math]I presume that the vector indicated as [math](-3,4)[/math] is [math]-3\hat i+4\hat j.[/math] This vector lies on the XY plane.Let a vector on the XY plane, orthogonal to this vector be [math]a\hat i+b\hat j.[/math][math]\Rightarrow\qquad \left(-3\hat i+4\hat j\right)\cdot \left(a\hat i+b\hat j\right)=0.[/math][math]\Rightarrow\qquad -3a+4b=0 \qquad\Rightarrow\qquad b=\frac{3a}{4}.[/math][math]\Rightarrow\qquad[/math] The vector [math]a\hat i+\frac{3a}{4}\hat j[/math] is orthogonal to the given vector for any real number [math]a.[/math]The magnitude of this vector is [math]\sqrt{a^2+\frac{9a^2}{16}}=\frac{5a}{4}.[/math][math]\Rightarrow\qquad[/math] The unit vectors orthogonal to the given vector is [math]\pm\frac{4}{5a}\left(a\hat i+\frac{3a}{4}\hat j\right)=\frac{4}{5}\hat i+\frac{3}{5}\hat j[/math] and [math]-\frac{4}{5}\hat i-\frac{3}{5}\hat j.[/math]All these vectors lie on the XY plane.The vector orthogonal to this vector has to be perpendicular to the XY plane.[math]\Rightarrow\qquad[/math] The unit vectors orthogonal to all these vectors is [math]\hat k[/math] and [math]-\hat k.[/math][math]\Rightarrow\qquad[/math] The the two unit vectors that are orthogonal to [math](-3,4)[/math] are any one of [math]\pm\left(\frac{4}{5}\hat i+\frac{3}{5}\hat j\right)[/math] and any one of [math]\pm\hat k.[/math]
Find two unit vectors orthogonal to both given vectors. 1, −1, 1 , 0, 4, 4?
Given : vectors. A = (1, −1, 1) B = (0, 4, 4) i j k AXB= 1 -1 1 0 4 4 AXB= -8i-4j+4k A unit vector is a vector whose length is 1. Therefore, the unit vector is => -8i-4j+4k/√(-8²+(-4)²+4²) => -8i-4j+4k/4√6 =>(-2/√6,-1/√6,1/√6) The second unit vector orthogonal to both <1,-1,1> and <0,4,4> would be (2/√6,1/√6,-1/√6) , which is the negative of the previous vector.
Find two unit vectors orthogonal to both i-j+k and 4j+4k.?
use the cross product (i-j+k)x(4j+4k) = -8i-4j+4k one unit vector=-8i-4j+4k/root 98, another is -16i-8j+8k/2root98, and so on good luck