Let two vectors x and y satisfy the following conditions - x - y-a- x-y - b-a- x -a-1 where a and b are non-zero perpendicular vectors- then

The correct option is A x = |a|^2b+a|a|^2 x⃗ +y⃗ =a⃗⟶ 1 x⃗×y⃗ =b⃗×a⃗⟶ 2 In 2 x⃗×a⃗ =1 x⃗×( a⃗ x⃗ #160; ) =b⃗ #160; ×a⃗ x⃗×a⃗ ...

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Byju's Answer

Standard XII

Mathematics

Applications of Dot Product

Let two vecto...

Question

Let two vectors $\to x$ and $\to y$ satisfy the following conditions : $\to x + \to y = \to a; \to x \times \to y = \to b \times \to a, \to x . \to a = 1$ where $\to a$ and $\to b$ are non-zero perpendicular vectors, then

\to x = \frac{| \to a |^{2} \to b + \to a}{| \to a |^{2}}

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\to x = \frac{| \to a |^{2} \to b - \to a}{| \to a |^{2}}

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\to y = \frac{| \to a |^{2} (\to a - \to b) - \to a}{| \to a |^{2}}

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\to y = \frac{\to a (\to a . \to y) - | \to a |^{2} \to b}{| \to a |^{2}}

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Solution

The correct option is A $\to x = \frac{| \to a |^{2} \to b + \to a}{| \to a |^{2}}$
$\to x + \to y = \to a ⟶ 1$
$\to x \times \to y = \to b \times \to a ⟶ 2$
In 2
$\to x \times \to a = 1$
$\to x \times (\to a - \to x) = \to b \times \to a$
$\to x \times \to a = \to b \times \to a$
(From equation 1 $\to y = \to a - \to x$ )
Now,
$\to a \times (\to x \times \to a) = \to a \times (\to b \times \to a)$
$(\to a \times \to a) \to x - (\to a \times \to x) \to a = (\to a \times \to a) \to b - (\to a \times \to b) \to a$
${| a |}^{2} \to x - \to a = {| a |}^{2} \to b - 0$ (because 1 vector)
$\to x = \frac{{| a |}^{2} \to b + \to a}{{| \to a |}^{2}}$

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Let two vectors →x and →y satisfy the following conditions : →x+→y=→a;→x×→y=→b×→a,→x.→a=1 where →a and →b are non-zero perpendicular vectors, then

Let two vectors $\to x$ and $\to y$ satisfy the following conditions : $\to x + \to y = \to a; \to x \times \to y = \to b \times \to a, \to x . \to a = 1$ where $\to a$ and $\to b$ are non-zero perpendicular vectors, then