Any point on a line has the coordinates (r+1,r−3,r2+4) where ‘r’ is any real number the d.r’s of the line are

p=(r+1,r−3,r2+4) ⇒r=0⇒A(1,−3,4)r=1⇒B(2,−2,2+4)DR’S of AB¯ is (1,1,2)

Any point on a line has the coordinates (r+1,r−3,r2+4) where ‘r’ is any real number the d.r’s of the line are

p=(r+1,r−3,r2+4) ⇒r=0⇒A(1,−3,4)r=1⇒B(2,−2,2+4)DR’S of AB¯ is (1,1,2)

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Any point on a line has the coordinates $(r + 1, r - 3, r \sqrt{2} + 4)$ where ‘r’ is any real number the d.r’s of the line are

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$\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}$

$\frac{1}{2}, \frac{- 1}{2}, \frac{1}{\sqrt{2}}$

$\frac{1}{2}, \frac{1}{2}, \frac{- 1}{\sqrt{2}}$

$1, 1, \sqrt{2}$

answer is D.

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Detailed Solution

$p = (r + 1, r - 3, r \sqrt{2} + 4) \Rightarrow r = 0 \Rightarrow A (1, - 3, 4)$
$r = 1 \Rightarrow B (2, - 2, \sqrt{2} + 4)$
DR’S of $\bar{A B}$ is $(1, 1, \sqrt{2})$

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Any point on a line has the coordinates (r+1,r−3,r2+4) where ‘r’ is any real number the d.r’s of the line are

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Any point on a line has the coordinates $(r + 1, r - 3, r \sqrt{2} + 4)$ where ‘r’ is any real number the d.r’s of the line are