The projection of the vector -k- on the line whose vector equation is r-3-t-2t-1-3tk- t being the scalar parameter is

Since r=3î 1ĵ+t(î+2ĵ+3k̂) So, the line is parallel to the vector (î+2ĵ+3k̂) Now, unit vector along the line is nbsp; (î+2ĵ+3k̂)√(14) ∴ The projection of nbs ...

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Standard XII

Mathematics

Cross Product of Two Vectors

The projectio...

Question

The projection of the vector $^i +^j +^k$ on the line whose vector equation is $\to r = (3 + t)^i + (2 t - 1)^j + 3 t^k, t$ being the scalar parameter is

\frac{6}{\sqrt{14}}

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\frac{1}{\sqrt{14}}

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10

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Solution

The correct option is A $\frac{6}{\sqrt{14}}$
Since $\to r = 3^i - 1^j + t (^i + 2^j + 3^k)$
So, the line is parallel to the vector $(^i + 2^j + 3^k)$
Now, unit vector along the line is
$\frac{(^i + 2^j + 3^k)}{\sqrt{14}}$
$∴$ The projection of $^i +^j +^k$ on the given line is $= (^i +^j +^k) \cdot \frac{(^i + 2^j + 3^k)}{\sqrt{14}} = \frac{6}{\sqrt{14}}$

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Cross Product of Two Vectors

Standard XII Mathematics

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The projection of the vector ^i+^j+^k on the line whose vector equation is →r=(3+t)^i+(2t−1)^j+3t^k, t being the scalar parameter is

The correct option is A 6√14Since →r=3^i−1^j+t(^i+2^j+3^k) So, the line is parallel to the vector (^i+2^j+3^k) Now, unit vector along the line is (^i+2^j+3^k)√14 ∴ The projection of ^i+^j+^k on the given line is =(^i+^j+^k)⋅(^i+2^j+3^k)√14=6√14

The projection of the vector $^i +^j +^k$ on the line whose vector equation is $\to r = (3 + t)^i + (2 t - 1)^j + 3 t^k, t$ being the scalar parameter is