Diketahui vektor a=(1, x, 2), b=(2,1,-1) dan panjang

5/2 akar(2)

Pembahasan

Diketahui vektor a=(1, x, 2) dan b=(2,1,-1).
Panjang proyeksi ortogonal vektor a pada b adalah |a · b| / |b|.

a · b = (1)(2) + (x)(1) + (2)(-1) = 2 + x - 2 = x

|b| = sqrt(2^2 + 1^2 + (-1)^2) = sqrt(4 + 1 + 1) = sqrt(6)

Panjang proyeksi ortogonal a pada b = |x| / sqrt(6).
Diketahui panjang proyeksi ortogonal adalah 1/3 akar(6), maka:
|x| / sqrt(6) = (1/3) * sqrt(6)
|x| = (1/3) * sqrt(6) * sqrt(6)
|x| = (1/3) * 6
|x| = 2
Jadi, x = 2 atau x = -2.

Nilai cos theta = (a · b) / (|a| |b|).
|a| = sqrt(1^2 + x^2 + 2^2) = sqrt(1 + x^2 + 4) = sqrt(5 + x^2)

Jika x = 2:
cos theta = 2 / (sqrt(5 + 2^2) * sqrt(6)) = 2 / (sqrt(9) * sqrt(6)) = 2 / (3 * sqrt(6)) = 2 / (3 sqrt(6))
Untuk mencari tan theta, kita bisa menggunakan identitas sin^2 theta + cos^2 theta = 1.
sin^2 theta = 1 - cos^2 theta = 1 - (2 / (3 sqrt(6)))^2 = 1 - (4 / (9 * 6)) = 1 - 4/54 = 1 - 2/27 = 25/27
sin theta = sqrt(25/27) = 5 / sqrt(27) = 5 / (3 sqrt(3))
tan theta = sin theta / cos theta = (5 / (3 sqrt(3))) / (2 / (3 sqrt(6))) = (5 / (3 sqrt(3))) * ((3 sqrt(6)) / 2) = (5 sqrt(6)) / (2 sqrt(3)) = (5 sqrt(2)) / 2

Jika x = -2:
cos theta = -2 / (sqrt(5 + (-2)^2) * sqrt(6)) = -2 / (sqrt(9) * sqrt(6)) = -2 / (3 sqrt(6))
sin^2 theta = 1 - (-2 / (3 sqrt(6)))^2 = 1 - (4 / 54) = 25/27
sin theta = 5 / (3 sqrt(3))
tan theta = sin theta / cos theta = (5 / (3 sqrt(3))) / (-2 / (3 sqrt(6))) = -(5 sqrt(6)) / (2 sqrt(3)) = -(5 sqrt(2)) / 2

Karena biasanya sudut antara dua vektor diambil yang lancip, maka kita ambil nilai positifnya.