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

Answer: (A)

$\sqrt{\frac{21}{2}}$

Answer 2

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):$\sqrt{\frac{21}{2}}$