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ArbitrLikvidat [17]
2 years ago
9

Let a⃗ =4ı^−2ȷ^, b⃗ =−3ı^+5ȷ^, and e⃗ =2a⃗ +3b⃗ . part a write vector e⃗ in component form.

Mathematics
1 answer:
ch4aika [34]2 years ago
4 0

We have been given that:

Vector \vec{a}=4\hat i-2\hat j and \vec{b}=-3\hat i+5\hat j.

Component form of a vector implies to write the vector with each individual components.

Now, our vector \vec e is given by

\vec e =2\vec a+3\vec b

Now, 2 \vec a=2(4\hat i- 2 \hat j)=8 \hat i -4 \hat j

and 3 \vec b= 3(-3 \hat i +5\hat j)=-9 \hat i + 15\hat j

Plugging the values  of 2 \vec a and 3\vec b, we get:

\vec e = 8 \hat i-4 \hat j-9 \hat i + 15 \hat j=- \hat i+11 \hat j

So, the vector 'e' in its component form is given as:

\vec e= -\hat i+11\hat j

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Step-by-step explanation:

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positive * negative = negative

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april bought some sports drinks and slices of pizza for her friend. she bought 3 more sports drinks than slices of pizza. if her
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s+p=21.25

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As she bought 3 more sports drinks than pizza slices so, s = p+3

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What is the product of (8.8 × 106)(5 × 102)?
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(8.8 × 106)(5 × 102)

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Show that the three points whose position vectors are 7j+ 10k,-i + 6j+6k and - 4i + +9j + 6k form an isosceles right
Novay_Z [31]

Answer:

AB = √18 , BC=√18 and CA =4

AB²+BC²  = CA² and AB=BC

ΔABC isosceles right  angled triangle.

Step-by-step explanation:

Given vectors are  7j+ 10k,-i + 6j+6k and - 4i + +9j + 6k

A( 0,7,10), B( -1,6,6) C(-4,9,6)

AB⁻ = OB-OA = -I+6j+6k-(7j+10k) = -I-j-4k

AB = \sqrt{1+1+16} = \sqrt{18}

BC = OC-OB = -4i+9j+6k-(-I+6j+6k) = -3i+3j

BC=\sqrt{9+9} =\sqrt{18}

CA = OA-OC = 7j+10k - (- 4i + +9j + 6k ) = 4i-2j+4k

CA = \sqrt{16+4+16} =\sqrt{36} =4

Since AB²+BC²  = CA²

And AB=BC

Therefore it follows that ΔABC is a right angled isosceles triangle



3 0
2 years ago
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