Let a- - 6i- - 3j- - 6k- 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 7i⃗ 2j⃗ 5k⃗ a⃗ = 6i⃗ 3j⃗ 6k⃗ d⃗ = i⃗ + j⃗ + k⃗ a⃗ = b⃗ + c⃗ ..... (1) b⃗ = λd⃗ and c⃗.d⃗ = 0 ...

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Applications of Dot Product

Let a⃗ = 6i...

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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\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

(\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

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

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

\to a = \to b + \to c

\to b is parallel to \to d and \to c is perpendicular to \to d .

\to c

\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

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

\to A = 2 \to i + \to k, \to B = \to i + \to j + \to k,

\to C = 4 \to i - 3 \to j + 7 \to k

\to R

\to R \times \to B = \to C \times \to B

\to R . \to A = 0

\to r = 3 \to i + 2 \to j - 5 \to k

\to a = 2 \to i - \to j + \to k

\to b = \to i + 3 \to j - 2 \to k

\to c = - 2 \to i + \to j - 3 \to k

\to r = λ \to a + μ \to b + ν \to c

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