Let a-6 i-3 j-6 k- and d-i-j-k- Suppose that a-b-c- whereb is parallel to d and c is perpendicular to d- Then C is A- 5 i -4 j - kB- 7 i -2 j -5 kC- 4 i -5 j - kD- 3 i -6 j -9 k

The correct option is B 7i 2j 5ka=6î 3ĵ 6k̂ d=î+ĵ+k̂ ∵ nbsp;b =λ nbsp;d= nbsp;λ nbsp;î+λ nbsp;ĵ+λ nbsp;k̂ and c.d= 0 Also, nbsp;a =b +c ⇒c=(6 λ ...

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Byju's Answer

Standard XII

Mathematics

Dot Product of Two Vectors

Let a⃗=6 i⃗-3...

Question

Let $\to a = 6 \to i - 3 \to j - 6 \to k$ and $\to d = \to i + \to j + \to k .$ Suppose that $\to a = \to b + \to c$ where $\to b is parallel to \to d and \to c is perpendicular to \to d .$ Then $\to c$ is

5 \to i - 4 \to j - \to k

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7 \to i - 2 \to j - 5 \to k

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4 \to i - 5 \to j + \to k

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3 \to i + 6 \to j - 9 \to k

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Solution

The correct option is B $7 \to i - 2 \to j - 5 \to k$
$\to a = 6^i - 3^j - 6^k$
$\to d =^i +^j +^k$
$∵ \to b = λ \to d = λ^i + λ^j + λ^k$
and $\to c . \to d = 0$
Also, $\to a = \to b + \to c$
$\Rightarrow \to c = (6 - λ)^i + (- 3 - λ)^j + (- 6 - λ)^k$
$∴ \to c . \to d = (6 - λ) + (- 3 - λ) + (- 6 - λ) = 0$
$\Rightarrow λ = - 1$
$∴ \to c = 7^i - 2^j - 5^k$

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Similar questions

Q. Let

\to a = 6 \to i - 3 \to j - 6 \to k

and

\to d = \to i + \to j + \to k

. Suppose that

\to a = \to b + \to c

where

\to b

is parallel to

\to d

and

\to c

is perpendicular to

\to d

. Then

\to c

is.

Q. If

\to a = \to i + \to j + \to k, \to a = 2 \to i + \to k, \to c = 2 \to i + \to j + \to k; \to d = \to i + \to j + 2 \to k

, then verify that

(\to a \times \to b) \times (\to c \times \to d) = [\to a \to b \to d] \to c - [\to a \to b \to c] \to d

Q. If

\to a = \to i - 2 \to j - 3 \to k, \to b = 2 \to i + \to j - \to k, \to c = \to i + 3 \to j - 2 \to k

then

(\to a \times \to b) \times \to c

Q. Let

\to a = \to i + \to j + \to k, \to c = \to j - \to k

. If

\to b

is a vector satisfying

\to a \times \to b = \to c

and

\to a . \to b = 3

, then

\to b

Q. Let

\to a

\to i

\to k

\to b

= x

\to i

\to j

+ (1-x)

\to k

and

\to c

= y

\to i

+ x

\to j

+ (1 + x - y)

\to k

. Then [

\to a

\to b

\to c

] depends on

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Let →a=6→i−3→j−6→k and →d=→i+→j+→k. Suppose that →a=→b+→c where →b is parallel to →d and →c is perpendicular to →d. Then →c is

The correct option is B 7→i−2→j−5→k→a=6^i−3^j−6^k →d=^i+^j+^k ∵→b=λ→d=λ^i+λ^j+λ^k and →c.→d=0 Also, →a=→b+→c ⇒→c=(6−λ)^i+(−3−λ)^j+(−6−λ)^k ∴→c.→d=(6−λ)+(−3−λ)+(−6−λ)=0 ⇒λ=−1 ∴→c=7^i−2^j−5^k

Let $\to a = 6 \to i - 3 \to j - 6 \to k$ and $\to d = \to i + \to j + \to k .$ Suppose that $\to a = \to b + \to c$ where $\to b is parallel to \to d and \to c is perpendicular to \to d .$ Then $\to c$ is