If a- and b- are not perpendicular to each other and r-a-b-a- r-c-0- then r- is equal to

The correct option is C #160;b (#160;b.#160;c)(#160;a.#160;c)#160;a Let r,a,b and c be vectors.It is given that r× a=b× a r× a (b× a) ...

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Standard XII

Mathematics

Position of a Line W.R.T Ellipse

If a⃗ and ...

Question

If $\to a$ and $\to b$ are not perpendicular to each other and $\to r \times \to a = \to b \times \to a, \to r . \to c = 0$ , then $\to r$ is equal to

\to a - \to c

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\to b + λ \to a

, for all scalars

λ

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\to b - \frac{(\to b . \to c)}{(\to a . \to c)} \to a

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\to a + \to c

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Solution

The correct option is C $\to b - \frac{(\to b . \to c)}{(\to a . \to c)} \to a$
Let $r, a, b$ and $c$ be vectors.
It is given that
$r \times a = b \times a$
$r \times a - (b \times a) = 0$
$(r - b) \times a = 0$
Hence $r - b$ is a vector parallel to vector $a$ .
$r = b + μ a$ ...(i)
It is given that $r . c = 0$ .
Hence $r$ vector is perpendicular to $c$ vector.
$(b + μ a) . c = 0$ ...(from i)
$b . c + μ (a . c) = 0$
$μ (a . c) = - b . c$
$μ = - \frac{b . c}{a . c}$
Hence $r = b - \frac{b . c}{a . c} a$

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Similar questions

Q. Let

\to a, \to b, \to c

be three vectors such that

\to a \neq 0

and

\to a \times \to b = 2 \to a \times \to c, | \to a | = | \to c | = 1, | \to b | = 4

and

| \to b \times \to c | = \sqrt{15}

. If

\to b - 2 \to c = λ \to a

, then

λ

is equal to

Q. If

\to a + 2 \to b + 3 \to c = 0

and

\to a \times \to b + \to b \times \to c + \to c \times \to a

is equal to

λ (\to b \times \to c)

then

λ

is equal to

Q. Suppose the three vectors

\to a, \to b, \to c

on a plane satisfy the condition that

| \to a | = | \to b | = | \to c | = | \to a + \to b | = 1; \to c

is perpendicular to

\to a

and

\to b . \to c > 0,

, then
(1)Find the angle formed by

2 \to a + \to b

and

\to b

(2)If the vector

\to c

is expressed as a linear combination

λ \to a + μ \to b

then find the ordered pair

(λ, μ)

Q. If

\to r \times \to a = \to b \times \to a; \to r \times \to b = \to a \times \to b; \to a \neq 0, \to b \neq 0, \to a \neq λ \to b; \to a

is not perpendicular to

\to b

, then

\to r =

Q. Assertion :If

\to r \times \to a = \to b, \to r . \to a = 1,

then

\to r = \to a + \to a \times \to b .

Reason: The solution of the

\to r \times \to a = \to b

\to r = \to a \times \to b

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If →a and →b are not perpendicular to each other and →r×→a=→b×→a, →r.→c=0, then →r is equal to

If $\to a$ and $\to b$ are not perpendicular to each other and $\to r \times \to a = \to b \times \to a, \to r . \to c = 0$ , then $\to r$ is equal to