What is the vector perpendicular to vector 4i+3j?
We want to find out a vector perpendicular to the vector [math]4\hat i+3\hat j.[/math]When two vectors are perpendicular to each other the angle between them is [math]90^o[/math] and consequently their dot product is zero.Let the vector [math]a\hat i+b\hat j+c\hat k[/math] be perpendicular to the vector [math]4\hat i+3\hat j,[/math] where [math]a,b,c[/math] are real numbers.[math]\Rightarrow\qquad \left(a\hat i+b\hat j+c\hat k\right)\cdot (4\hat i+3\hat j)=0.[/math][math]\Rightarrow\qquad4a+3b=0 \qquad\Rightarrow\qquad b=-\frac{4}{3}a.[/math][math]\Rightarrow\qquad[/math] Vectors perpendicular to the vector [math]4\hat i+3\hat j[/math] are of the type[math]\qquad\qquad a\hat i-\frac{4}{3}a\hat j+c\hat k,\qquad a,c\,\,\in\,\, R.[/math]
What is the vector perpendicular to vector 4i-3j?
Talking about two perpendicular vectors,The dot productof both is zero. Always.a•b= ab cosαSo, the vector is 4i-3jBy the above relation we get,(4i-3j).(4i-3j).(ai+bj)=04a-3b=04a=3ba/b=3/4Therefore the vector perpendicular to given vector is 3i+4j.
If A =2i^ +4j^, what is the unit vector perpendicular to A?
A shortcut I often use to find a vector perpendicular to another vector, is to use the fact that [math]a\hat{i}+b\hat{j}[/math] is perpendicular to [math]b\hat{i}-a\hat{j}[/math], because their dot product is [math]ab-ba=0[/math].So a vector perpendicular to [math]2\hat{i}+4\hat{i}[/math] is [math]4\hat{i}-2\hat{i}[/math].Now all that remains is to normalize this vector, giving us[math]\dfrac{2\hat{i}-\hat{i}}{\sqrt{5}}[/math]
If a1i^+a2j^ is a unit vector perpendicular to 4i^-3j^, what is a1 and a2?
If these 2 vectors are perpendicular, their Dot product must be Zero.Which means 4(a1)-3(a2)=0or we can write (a1) = 3(a2)/4.—————-(1)Also, since (a1)i+(a2)j is a unit vector, it means - sqrt[(a1*a1)+(a2*a2)] = 1, {sqrt = square root}squaring both sides, we get (a1)*(a1)+(a2)*(a2) = 1———————-(2)Putting value of (a1) in eqn(2), we get - [3(a2)/4]*[3(a2)/4] + (a2)*(a2) = 1.Solving further, 25(a2)*(a2) = 16Take square root both sides, +-5(a2) = +-4It gives 2 values of (a2) which are +4/5 & -4/5.Correspondingly (a1) = 3(a2)/4 becomes +3/5 & -3/5.Hence we obtain 2 solutions for the question, (a1) = +3/5, (a2) = +4/5 and (a1) = -3/5, (a2) = - 4/5.
What is the unit vector parallel to I+j+k?
The vector (i + j + k) is (1, 1, 1) which has a magnitude R ofR = sqrt(1^2 + 1^2 + 1^2) = sqrt(3) = 1.732so the unit vector u parallel to (1, 1, 1) is found by dividing this magnitude by 1.732u = (1, 1, 1)/1.732 = (1/1.732, 1/1.732, 1/1.732) = (0.577, 0.577, 0.577)
If a1i+a2j is a unit vector then what is a1+a2?
Let unit vector is v = (a1 )i + (a2) j(a1)^2 + (a2)^2 = 1So we can relate both unknown with cosx and sinxa1 + a2 = cosx + sinxAnd cosx + sinx lies between -√2 & +√2It is range of values and not an absolute value
F() is a function defined on vectors of size K with natural numbers: F (a1,a2,a3,..., aK) = (a1*a2*a3*...*aK) + (a1+a2+a3+...+aK). Is it a strict one-to-one function (i.e. for every different vector I get different values)?
For [math]K = 2[/math], [math]F(1, 5) = F(2, 3)[/math], so the answer is no.
Urgent math question perpendicular unit vectors 2D plane?
Let ai + bj be a unit vector perpendicular to 4i - 3j. Since the vectors are perpendicular the scalar product is 0. (ai + bj) . (4i - 3j) = 0 4a - 3b = 0 Also, since ai + bj is a unit vector: a² + b² = 1 Substitute b = 4a/3 into a² + b² = 1. a² + 16a²/9 = 1 25a² = 9 a = 3/5 or -3/5 Substitute a = 3/5 and a = -3/5 into 4a - 3b = 0 12/5 - 3b = 0 or -12/5 - 3b = 0 b = 4/5 or b = -4/5 So there are two possible answers: (3/5)i + (4/5)j and the inverse of that.
If you apply BODMAS then2+3+4×6=2+3+24=29
If a and b are two non-parallel vectors satisfying |a|=|b|, then the vector (a+b) ×(a×b) is parallel to what?
It will be parallel to a vector which is parallel to both a+b and a×bI think we can't simplify this question more than thisu can try taking a=3i+4j and b=4i-3jwhich are perpendicular as u said a and b are non parallelthis might be the easiest case to solve