R vector is equal to AC vector minus TK vector is the equation of
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Step-by-step explanation:
let a=2i+2j-2k, b=5i+yj+k and c=-i+2j+2k
If a,b and c are coplanar, then
[a b c]=0 i.e. scalar triple product=0
=> (a x b) . c=0
or (c x a) . b=0
we have,
c x a = 6j-6k
therefore, (6j-6k).(5i+yj+k)=0
=>6y-6=0
=>y=1
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If 2i - 6j + k and 5i + 2j + αk are perpendicular vectors, then what is the value of α?
How do I find the vector r which is perpendicular to i-2j+5k and 2i+3j-k and r (2i+j+k)?
What is the number of a unit vector perpendicular to the following vectors: 2i+2j-k and 6i-3j+2k?
If vector A is equal to 5i+2j - 3k and vector B is equal to 3i -2j+2k, then what is the angle between the two vectors?
What is the answer of (-3i+2k) (2i+2j+2k)?
Let,
a⃗ =2i^+2j^−2k^
b⃗ =5i^+yj^+k^
c⃗ =−i^+2j^+2k^
If a⃗ ,b⃗ ,c⃗ are coplanar, then their scalar triple product should be zero.
[a⃗ b⃗ c⃗ ]=0
That implies,
(a⃗ ×b⃗ ).c⃗ =0
By cyclic rotation,
(c⃗ ×a⃗ ).b⃗ =0
We have,
c⃗ ×a⃗ =6j^−6k^
Therefore,
(6j^−6k^).(5i^+yj^+k^)=0
6y−6=0
y=1
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The solution by Czaee Shefali Kolekar is absolutely correct. So I don't have to answer again.:)
But I would like to tell you how scalar product works in this case.
You have three different vectors which are said to be coplanar. Coplanar, as its name suggests, means lying on the same plane. Thus, they will have a common normal ( like for all the lines on xy plane, we have z-axis as common normal).
Now, in case of three vectors, we can find a normal for two know vectors by simply taking their cross product.
a*b=ñ ( normal )
For all the lines on plane formed by a and b vectors, ñ will be normal. Also, scalar product of two perpendicular vector is zero. In that way, the scalar product of ñ and the third vector ( which has unknown term ) will be zero.
Thus, we can easily find the unknown term.
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