Vector Cross Product: Calculating -i x i = 0?

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The discussion centers on calculating the cross product of -i x i, which simplifies to -(i x i). It is established that the cross product of any vector with itself is zero, leading to the conclusion that -i x i equals zero. The conversation then shifts to the projection of vector u (-i + 2j) onto vector v (i + 2j), with one participant initially misidentifying the operation as a cross product instead of a dot product. The correct formula for the projection is confirmed, emphasizing that the cross product is not relevant in this context. Ultimately, the participants clarify their understanding of vector operations and the distinction between dot and cross products.
Ry122
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What is the cross product of -i x i? Is it negative 1 or is still just 0?
 
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Note that for any 2 vectors a,b: -a x b = -(a x b). This reduces the problem to -(i x i).

Now what is the vector product of a vector with itself?
 
well can you tell me how the projection of u on to v
where
u=-i+2j and v=i+2j is
v=(3/5)i+(6/5)j ?
The answer i got was (4/5)i + (8/5)j
i used the equation
w=v.((u.v)/(modulusv^2))
 
Ry122 said:
i used the equation
w=v.((u.v)/(modulusv^2))
This equation for the projection is

\mathbf w = \mathbf v \frac {\mathbf u \cdot \mathbf v}{v^2}

Note well: The cross product is not involved when you compute the projection this way.
 
That's the same equation that I gave. yeah i realized my mistake after posting, i should have said dot product, not cross product.
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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