Question

The length of the perpendicular from the point (1, –2, 5) on the line passing through (1, 2, 4) and parallel to the line x + y – z = 0 = x – 2y + 3z – 5 is

• A. \(\sqrt{\frac{21}{2}}\)
• B. \(\sqrt{\frac{9}{2}}\)
• C. \(\sqrt{\frac{73}{2}}\)
• D. 1

Answer 1

The Correct Option is A

Fig.

The line x + y – z = 0 = x – 2y + 3z – 5 is parallel to the vector
\(\vec{b} = \begin{vmatrix}     \hat{i} & \hat{j} & \hat{k} \\     1 & 1 & -1 \\     1 & -2& 3 \end{vmatrix} = (1, 4, -3)\)
Equation of line through P(1, 2, 4) and parallel to
\(\frac{x−1}{1}=\frac{y−2}{−4}=\frac{z−4}{−3}\)
\(Let\ N ≡ (λ+1,−4λ+2,−3λ+4)\)
\(\stackrel{→}{QN}=(λ,−4λ+4,−3λ–1)\)
\(\stackrel{→}{QN} is\ perpendicular\ to\ \vec{b}\)
⇒ (λ,−4λ+4,−3λ–1)⋅(1,4,−3)=0
\(⇒λ=\frac{1}{2}\)
Hence \(\stackrel{→}{QN}=(\frac{1}{2},2,\frac{−5}{2})\)
and \(|\stackrel{→}{QN}|=\sqrt{\frac{21}{2}}\)
So, the correct option is (A):\(\sqrt{\frac{21}{2}}\)