Let- a- - - - 2 - -k-b- - - - - - k- -c- - - - - - k- A vector coplanar to a- and b- has a projection along c- of magnitude 1-3- then the vector is4- - - - 4k-4- - - - 4k2- - - - k-None of these

The correct option is A 4î ĵ + 4k̂Let vector r⃗ be coplanar to a⃗ and b⃗. ∴r⃗ = a⃗ + t b⃗ ⇒r⃗ = (î + 2ĵ + k̂) + t (î ĵ + k̂) = î (1 + t) + ĵ (2 t) + k̂ (1 ...

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Byju's Answer

Standard VII

Mathematics

Converse of Pythagoras Theorem

Let, a⃗ = î +...

Question

Let, $\to a =^i + 2^j +^k, \to b =^i -^j +^k, \to c =^i +^j -^k .$
A vector coplanar to $\to a$ and $\to b$ has a projection along $\to c$ of magnitude $\frac{1}{\sqrt{3}}$ , then the vector is

4^i -^j + 4^k

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4^i +^j - 4 k

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2^i +^j +^k

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None of these

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Solution

The correct option is A $4^i -^j + 4^k$
Let vector $\to r$ be coplanar to $\to a$ and $\to b . ∴ \to r = \to a + t \to b \Rightarrow \to r = (^i + 2^j +^k) + t (^i -^j +^k) =^i (1 + t) +^j (2 - t) +^k (1 + t)$
The projection of $\to r$ on $\to c = \frac{1}{\sqrt{3} .}$ [given]
$\Rightarrow \frac{\to r . \to c}{| \to c |} = \frac{1}{\sqrt{3}} \Rightarrow \frac{| 1. (1 + t) + 1. (2 - t) - 1. (1 + t) |}{\sqrt{3}} = \frac{1}{\sqrt{3}}$
$\Rightarrow (2 - t) = \pm 1 \Rightarrow t = 1 o r 3$
When, t = 1, we have $\to r = 2^i +^j + 2^k$
When, t = 3, we have $\to r = 4^i -^j + 4^k$

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Let, →a=^i+2^j+^k, →b=^i−^j+^k, →c=^i+^j−^k. A vector coplanar to →a and →b has a projection along →c of magnitude 1√3, then the vector is

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A vector coplanar to $\to a$ and $\to b$ has a projection along $\to c$ of magnitude $\frac{1}{\sqrt{3}}$ , then the vector is