find the vector equation of a line passing through the point with position vector 2i-3j-5k and perpendicular to the plane r.(6i-3j+5k)+2=0
Answers
We have to find the vector equation of a line passing through the point with position vector (2i - 3j + 5k) and perpendicular to the plane, r.(6i - 3j + 5k) + 2 = 0
Solution : equation of line passing through a and perpendicular to plane n is given by, r = a + λ n
⇒r = (2i - 3j - 5k) + λ(6i - 3j + 5k) , this is required equation of line.
Now point of intersection of line with the plane be (2 + 6λ, -3 - 3λ, -5 + 5λ). This point will satisfy plane r(6i - 3j + 5k) + 2 = 0
⇒6(2 + 6λ) - 3(-3 - 3λ) + 5(-5 + 5λ) + 2 = 0
⇒12 + 36λ + 9 + 9λ - 25 + 25λ + 2 = 0
⇒-2 + 70λ = 0
⇒λ = 1/35
Now putting λ = 1/35 into point of intersection,we get (76/35, -108/35, -34/7).
also read similar questions : Find the cross product of A = 2i + 3j + k and B = i -j + 2k. *
7i -3j -5k
7i + 3j + 5k
5i -7j +5k
5i +7j +5k
https://brainly.in/question/22266826
find a vector whose magnitude is 12 and which is perpendicular to each of the vectors, A=2i^+3j^-2k^ and B=6i^+5j^-2k^
https://brainly.in/question/22740622
Step-by-step explanation:
Cytoplasm is a thick solution that fills each cell and is enclosed by the cell membrane. It is mainly composed of water, salts, and proteins. ... All of the organelles in eukaryotic cells, such as the nucleus, endoplasmic reticulum, and mitochondria, are located in the cytoplasm.