If A - B - C and D are the points with position vectors -k- 2 -3 k- and 3 -2 -k- respectively- then find the projection ofA B along C D-

Here, O⃗A⃗=î+ĵ k̂,O⃗B⃗=2î ĵ+3k̂,O⃗C⃗=2î 3k̂ and O⃗D⃗=3î 2ĵ+k̂ ∴ A⃗B⃗=O⃗B⃗ O⃗A⃗=(2 1)î+( 1 1)ĵ+(3+1)k̂ =î 2ĵ+4k̂ and C⃗D⃗=O⃗D⃗ O⃗C⃗=(3 2)î+( 2 0)ĵ+(3+1)k̂ ...

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Byju's Answer

Standard XII

Physics

Vectors and Its Types

If A , B , C ...

Question

If A, B, C and D are the points with position vectors $^i -^j +^k, 2^i -^j + 3^k a n d 3^i - 2^j +^k$ respectively, then find the projection of $\to A B a l o n g \to C D$ .

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Solution

$Here, \to O A =^i +^j -^k, \to O B = 2^i -^j + 3^k, \to O C = 2^i - 3^k a n d \to O D = 3^i - 2^j +^k ∴ \to A B = \to O B - \to O A = (2 - 1)^i + (- 1 - 1)^j + (3 + 1)^k =^i - 2^j + 4^k and \to C D = \to O D - \to O C = (3 - 2)^i + (- 2 - 0)^j + (3 + 1)^k =^i - 2^j + 4^k So, the projection of \to A B a l o n g \to C D = \to A B . \frac{\to C D}{| \to C D |} = \frac{(^i - 2^j + 4^k) . (^i - 2^j + 4^k)}{\sqrt{1^{2} + 2^{2} + 4^{2}}} = \frac{1 + 4 + 16}{\sqrt{21}} = \frac{21}{\sqrt{21}} = \sqrt{21} u n i t s$

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If A, B, C and D are the points with position vectors ^i−^j+^k,2^i−^j+3^k and 3^i−2^j+^k respectively, then find the projection of →AB along →CD.

If A, B, C and D are the points with position vectors $^i -^j +^k, 2^i -^j + 3^k a n d 3^i - 2^j +^k$ respectively, then find the projection of $\to A B a l o n g \to C D$ .