Ned to solved some vector algebra proble.

mathseeker

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Show That (a ⃗-b ⃗)×(a ⃗+b ⃗)=2a ⃗×b ⃗
Show That a ⃗ ×(b ⃗+c ⃗)+b ⃗ ×(c ⃗+a ⃗)+c ⃗ ×(a ⃗+b ⃗)=0


Find the scalar m such that the scalar product of i+j+k with unit vector parallel to the sum of 2i+4j-5k and mi+2j+3k is equal to unity.

show that a=3i-2j+k, b=i-3j+5k, c=2i+j-4k form a right angled triangle
 
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ksdhart2

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Well, first off, it looks like whatever you tried to copy and paste didn't come through as intended. In the first two problems, I see a bunch of white boxes throughout. Please reply with a corrected version of these problem statements. Additionally, please also comply with the rules as laid out in the Read Before Postinghttps://www.freemathhelp.com/forum/threads/54004-Read-Before-Posting thread that's stickied at the top of each sub-forum (you did read it, right? ;)) and share with us any and all work you've done on this problems, even the parts you know for sure are wrong. Thank you.
 

mathseeker

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first two problems are

Show That

(vector a- vector b) X (vector a +vector b) = 2 vector a X vector b

and

Show That

Vector a X (Vector b + Vector C) + Vector b X (Vector c + Vector a) + Vector C X (Vector a + Vector b) = 0
 

MarkFL

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Suppose:

\(\displaystyle \textbf{a}=a_1\textbf{i}+a_2\textbf{j}+ a_3\textbf{k}\)

\(\displaystyle \textbf{b}=b_1\textbf{i}+b_2\textbf{j}+ b_3\textbf{k}\)

Then, we have:

\(\displaystyle \textbf{a}\times\textbf{b}= \left(a_2b_3-a_3b_2\right)\textbf{i}+ \left(a_3b_1-a_1b_3\right)\textbf{j}+ \left(a_1b_2-a_2b_1\right)\textbf{k}\)

Now, we are asked to verify the identity:

\(\displaystyle (\textbf{a}-\textbf{b}) \times(\textbf{a}+\textbf{b})= 2(\textbf{a}\times\textbf{b})\)

Can you compute the vector difference and sum needed on the left side?
 

HallsofIvy

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For the second problem, "Show that \(\displaystyle \vec{a}\times (\vec{b}+ \vec{c})+ \vec{b}\times(\vec{c}+ \vec{a})+ \vec{c}\times (\vec{b}+ \vec{a})= \vec{0}\)", you could actually work out each term using \(\displaystyle \vec{a}= a_1\vec{i}+ a_2\vec{j}+ a_3\vec{k}\), \(\displaystyle \vec{b}= b_1\vec{i}+ b_2\vec{j}+ b_3\vec{k}\), \(\displaystyle \vec{a}= c_1\vec{i}+ c_2\vec{j}+ c_3\vec{k}\) and \(\displaystyle \vec{c}= c_1\vec{i}+ c_2\vec{j}+ c_3\vec{k}\) and the formula for the cross product MarkFL gave. But simpler would be to use the fact that "the cross product distributes over addition", \(\displaystyle \vec{a}\times (\vec{b}+ \vec{c})= \vec{a}\times\vec{b}+ \vec{a}\times \vec{c}\), and the fact that the cross product is "anti-commutative", \(\displaystyle \vec{a}\times\vec{b}= -\vec{b}\times\vec{a}\). If you have not already learned those you can prove them, by using the formula MarkFL gave, once then use them repeatedly.

The third and fourth problems are easier in that they do not require the cross product. Can we presume at leas that you do know what a "scalar product" is and how to show that two vectors are parallel? Or that you know how to find the length of a vector and that a triangle is a right triangle if and only if the lengths of its sides satisfy the "Pythagorean Theorem"?
 
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mathseeker

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The third and fourth problems are easier in that they do not require the cross product. Can we presume at leas that you do know what a "scalar product" is and how to show that two vectors are parallel? Or that you know how to find the length of a vector and that a triangle is a right triangle if and only if the lengths of its sides satisfy the "Pythagorean Theorem"?
BUT HERE SAYS THAT is equal to unity. WHAT'S THAT MEANS?
 

mathseeker

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Find the scalar m such that the scalar product of i+j+k with unit vector parallel to

Find the scalar m such that the scalar product of i+j+k with unit vector parallel to the sum of 2i+4j-5k and mi+2j+3k is equal to unity.
 

Subhotosh Khan

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