Consider the vector equation for x- y- given by x-x-y-a- and y-x-y-b- where a- b- are fixed vectors- If solution for x- is x -1 a - m b - n a - b then identify the correct statement- - a - a - b - b A- If a -1-2- b - a - b -0 then 1-3 m -3 n -2B- If a -1-2- b - a - b -0 then 1-3 m - n -1C- If a -1- b -2 and a - b -1 then 1-3 m - n -2D- If a -1- b -2 and a - b -1 then 1-3 m -3 n -1

The correct options are B If a=1/2=b, a.b=0 then l+3m n=1 D If a=1, b=2 and a.b=1 then l 3m 3n=1x=l a + m b+n(a×b) …… (1) y=(l 1)a + (m+1) b+n(a×b) …… (2) Put ...

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Standard XII

Physics

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Question

Consider the vector equation for $\to x$ , $\to y$ given by $\to x + \to x \times \to y = \to a$ and $\to y + \to x \times \to y = \to b$ where $\to a$ , $\to b$ are fixed vectors. If solution for $\to x$ is $\to x = l \to a + m \to b + n (\to a \times \to b)$ then identify the correct statement? $(| \to a | = a, | \to b | = b)$

a = \frac{1}{2} = b, \to a . \to b = 0

then

l - 3 m - 3 n = 2

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a = \frac{1}{2} = b, \to a . \to b = 0

then

l + 3 m - n = 1

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a = 1, b = 2

and

\to a . \to b = 1

then

l + 3 m - n = 2

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a = 1, b = 2

and

\to a . \to b = 1

then

l - 3 m - 3 n = 1

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

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Solution

The correct options are
B If $a = \frac{1}{2} = b, \to a . \to b = 0$ then $l + 3 m - n = 1$
D If $a = 1, b = 2$ and $\to a . \to b = 1$ then $l - 3 m - 3 n = 1$
$\to x = l \to a + m \to b + n (\to a \times \to b) \dots \dots (1)$

$\to y = (l - 1) \to a + (m + 1) \to b + n (\to a \times \to b) \dots \dots (2)$

Putting (1) and (2) in $\to x + \to x \times \to y = \to a$

$(l + m + n) (\to a \times \to b) + [n (\to a . \to b - | \to b |^{2}) + l - 1] \to a + [n (\to a . \to b - | \to a |^{2}) + m] \to b = 0$

$\Rightarrow l + m + n = 0$

$n (\to a . \to b - | \to b |^{2}) + l = 1$

$n (\to a . \to b - | \to a |) + m = 0$

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Consider the vector equation for →x, →y given by →x+→x×→y=→a and →y+→x×→y=→b where →a, →b are fixed vectors. If solution for →x is →x=l→a+m→b+n(→a×→b) then identify the correct statement? (|→a|=a,|→b|=b)