Given \(\vec A=2\hat i+\hat k\)
\(\vec B\) = \(\hat i+3\hat j-9\hat k\)
\(\vec V_2\) is parallel to \(\vec A\).
\(\therefore\) \(\vec V_2\) = \(\lambda(2\hat i+\hat k)\)
\(\vec V_1\) is perpendicular to \(\vec B\)
\(\vec V_1.\vec B=0\)----(1)
Let \(\vec V_1\) = x\(\hat i\) + y\(\hat j\) + z\(\hat k\)
Given that \(\vec B=\) V1 - 2V2
⇒ (\(\hat i+3\hat j-9\hat k\)) = (x\(\hat i\) + y\(\hat j\) + z\(\hat k\)) - 2\(\lambda\)(2\(\hat i\) + \(\hat k\))
⇒ \(\hat i+3\hat j-9\hat k\) = (x - h\(\lambda\))\(\hat i\) + y\(\hat j\) + (z - 2\(\lambda\))\(\hat k\)
\(\therefore\) x - 4\(\lambda\) = 1, y = 3
z - 2\(\lambda\) = -9------(2)
From (1), \(\vec V_1,\vec B=0\)
\((x\hat i+y\hat j+z\hat k).(\hat i+3\hat j-9\hat k)=0\)
⇒ x + 3y - 9z = 0
⇒ 1 + 4\(\lambda\) + 9 - 9(-9 + 2\(\lambda\)) = 0 (From(2))
⇒ 10 + 4\(\lambda\) + 81 + 18\(\lambda\) = 0
⇒ 14\(\lambda\) = 91
⇒ \(\lambda\) = 91/14 = 13/2
\(\therefore\) x = 1 + 4\(\lambda\) = 1 + 4 x 13/2 = -9 + 13 = 4
\(\therefore\) V1 = 27\(\hat i\) + 3\(\hat j\) + 4\(\hat k\)
V2 = 13/2(2\(\hat i\) + \(\hat k\)) = 13\(\hat i\) + 13/2 \(\hat k\)