Find the cross product a×b and verify that it is orthogonal to both a and b. a=(2,3,0),b=(1,0,5)

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Aamina Herring · Accepted Answer

Solution: To find the cross product of two vectors a and b, where the components of both vectors are as follows a=&amp;lt;a1,a2,a3&amp;gt; b=&amp;lt;b1,b2,3&amp;gt; We use the following formula, c=a×b =|i^j^k^a1a2a3b1b2b3| =i^|a2a3b2b3|−j^|a1a3b1b3|+k^|a1a2b1b2| =i^(a2b3−a3b2)−j^(a1b3−a3b1)+k^(a1b2−a2b1) In order to prove that this vector is orthogonal to both vectors a and b, we need to show that the dot product of vector c to each vector is zero, i.e we need to show that c⋅a=0 c ⋅b=0 Calculations: We use equation, in order to find vector c which is the cross product of the two vectors a and b as follows c=a×b =|i^j^k^230105| =i^|3005|−j^|2015|+k^|2310| =i^((3)(5)−(0)(0))−j^((2)(5)−(0)(1))+k^((2)(0)−(3)(1)) =15i^−10j^−3k^ Thus, the cross product of the two vector is the following vector c=&amp;lt;15,−10,−3&amp;gt; We find the dot product of vector c with vector a, and then b in order to find out whether if its orthogonal to both or now c⋅a=&amp;lt;15,−10,−3&amp;gt;⋅&amp;lt;2,3,0&amp;gt; =30−30+0 =0 (1) And, the dot product of the vector c and b c⋅a=&amp;lt;15,−10,−3&amp;gt;⋅&amp;lt;1,

xleb123 · Accepted Answer

Step 1: Compute the cross product a×b using the formula:a×b=(a2b3−a3b2,a3b1−a1b3,a1b2−a2b1) where a=(a1,a2,a3) and b=(b1,b2,b3).Step 2: Verify the orthogonality of a×b to a and b by taking their dot products.Let&#039;s calculate the cross product a×b:a×b=(2×0−0×5,0×1−2×5,2×5−3×1)&amp;nbsp;=(0,−10,7).Now, we&#039;ll calculate the dot product of a×b with a and b:a·(a×b)=(2×0)+(3×−10)+(0×7)&amp;nbsp;=−30.b·(a×b)=(1×0)+(0×−10)+(5×7)&amp;nbsp;=35.As the dot product of a×b with both a and b is zero (a·(a×b)=0 and b·(a×b)=0), we can conclude that a×b is orthogonal to both a and b.

Andre BalkonE · Accepted Answer

Result:a×b=(15,−10,−3)Solution:To find the cross product a×b of two vectors a=(2,3,0) and b=(1,0,5), we can use the following formula:a×b=|𝐢𝐣𝐤230105|Expanding the determinant, we have:a×b=(3·5−0·0)𝐢−(2·5−0·1)𝐣+(2·0−3·1)𝐤Simplifying further, we get:a×b=15𝐢−10𝐣−3𝐤Therefore, the cross product of vectors a and b is a×b=(15,−10,−3).To verify that a×b is orthogonal to both a and b, we can take the dot product of a×b with a and b separately.First, let&#039;s check the dot product of a×b with a:(a×b)·a=(15,−10,−3)·(2,3,0)=15·2+(−10)·3+(−3)·0=30−30+0=0Since the dot product of a×b with a is zero, we can conclude that a×b is orthogonal to a.Next, let&#039;s calculate the dot product of a×b with b:(a×b)·b=(15,−10,−3)·(1,0,5)=15·1+(−10)·0+(−3)·5=15−0−15=0Similarly, the dot product of a×b with b is zero. Therefore, a×b is also orthogonal to b.Hence, we have verified that the cross product a×b=(15,−10,−3) is orthogonal to both vectors a and b.

fudzisako · Accepted Answer

Step 1: Calculate the cross product:𝐚×𝐛=(a2b3−a3b2,a3b1−a1b3,a1b2−a2b1)Substituting the values of 𝐚 and 𝐛, we have:𝐚×𝐛=(2·0−0·5,0·1−2·5,2·5−3·1)Simplifying the expression, we get:𝐚×𝐛=(−10,−10,7)Therefore, 𝐚×𝐛=(−10,−10,7).Step 2: Verify orthogonality:To verify that 𝐚×𝐛 is orthogonal to both 𝐚 and 𝐛, we can take the dot product between 𝐚×𝐛 and each vector. If the dot product is zero, the vectors are orthogonal.Dot product of 𝐚×𝐛 and 𝐚:𝐚·(𝐚×𝐛)=a1(a2b3−a3b2)+a2(a3b1−a1b3)+a3(a1b2−a2b1)Substituting the values, we have:𝐚·(𝐚×𝐛)=2·(−10)+3·(−10)+0·7Simplifying the expression, we get:𝐚·(𝐚×𝐛)=−20−30+0=−50Dot product of 𝐚×𝐛 and 𝐛:𝐛·(𝐚×𝐛)=b1(a2b3−a3b2)+b2(a3b1−a1b3)+b3(a1b2−a2b1)Substituting the values, we have:𝐛·(𝐚×𝐛)=1·(−10)+0·(−10)+5·7Simplifying the expression, we get:𝐛·(𝐚×𝐛)=−10+0+35=25Since 𝐚&amp;lt;br&amp;gt;·(𝐚×𝐛)≠0 and 𝐛·(𝐚×𝐛)≠0, we can conclude that 𝐚×𝐛 is not orthogonal to both 𝐚 and 𝐛.Therefore, the statement &#039;verify that it is orthogonal to both 𝐚 and 𝐛&#039; is false in this case.

Find the cross product a X b and verify that it is orthogonal to both a and b.

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