Distance in units of a point with position vector -2-k- from the line given by L-r-2- -3k- -k- is-

We know d=|(p a)×c||c| Given: nbsp;L:r=2î+ ĵ+3k̂+λ (î+ĵ+k̂), p=î+2ĵ+k̂ Comparing with r= a+λc, we have (a)=2î+ ĵ+3k̂ and c=î+ĵ+k̂ Here, (p a)= î+ĵ 2 k̂ Now, ...

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Byju's Answer

Standard XII

Mathematics

Perpendicular Distance of a Point from a Line

Distance in u...

Question

Distance $(in units)$ of a point with position vector $^i + 2^j +^k$ from the line given by $L : \to r = 2^i +^j + 3^k + λ (^i +^j +^k)$ is:

\sqrt{\frac{14}{3}}

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\frac{\sqrt{7}}{3}

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\frac{\sqrt{14}}{3}

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\sqrt{\frac{7}{3}}

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Solution

The correct option is A $\sqrt{\frac{14}{3}}$
We know
$d = \frac{| (\to p - \to a) \times \to c |}{| \to c |}$
Given: $L : \to r = 2^i +^j + 3^k + λ (^i +^j +^k), \to p =^i + 2^j +^k$
Comparing with $\to r = \to a + λ \to c,$ we have $(\to a) = 2^i +^j + 3^k$ and $\to c =^i +^j +^k$
Here, $(\to p - \to a) = -^i +^j - 2^k$
Now, $(\to p - \to a) \times \to c = ∣ ∣ ∣ ∣ ∣ \begin{matrix} ^i & ^j & ^k - 1 & 1 & - 2 1 & 1 & 1 \end{matrix} ∣ ∣ ∣ ∣ ∣ = 3^i -^j - 2^k$
$\Rightarrow | \to c | = \sqrt{3}$
$d = \frac{| (\to p - \to a) \times \to c |}{| \to c |} = \frac{| 3^i -^j - 2^k |}{\sqrt{3}}$
Thus, $d = \sqrt{\frac{14}{3}}$ units

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Distance (in units) of a point with position vector ^i+2^j+^k from the line given by L:→r=2^i+^j+3^k+λ(^i+^j+^k) is:

Distance $(in units)$ of a point with position vector $^i + 2^j +^k$ from the line given by $L : \to r = 2^i +^j + 3^k + λ (^i +^j +^k)$ is: