Let a- - - - - b- - j- - k- c- - k- - - If d- is a unit vector such that a- d- - 0 - -b- c- d- then d- equals

Let d⃗ = xî + y ĵ + z k̂ where, x^2 + y^2 + z^2 = 1 ...(i) [∵ d being unit vector] Since, a⃗. d⃗ = 0 ⇒ x y = 0 ⇒ x = y ...(ii) Als ...

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Scalar Triple Product

Let a⃗ = î - ...

Question

Let $\to a =^i -^j, \to b = \to j -^k, \to c =^k -^i . I f \to d$ is a unit vector such that $\to a . \to d = 0 = [\to b \to c \to d],$ then $\to d$ equals

\pm \frac{^i +^j - 2^k}{\sqrt{6}}

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\pm \frac{^i +^j -^k}{\sqrt{3}}

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\pm \frac{^i +^j +^k}{\sqrt{3}}

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\pm^k

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Solution

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\to a =^i -^j, \to b = \to j -^k, \to c =^k -^i . I f \to d

\to a . \to d = 0 = [\to b \to c \to d],

\to d

\to a =^i +^j +^k, \to b =^i +^j -^k

\to c =^i -^j +^k

\to d =^i -^j -^k

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

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

\to c = -^i +^j - 4^k

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

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

\to a =^i +^j +^k

\to b =^i +^j -^k

\to c =^i -^j +^k

\to d =^i -^j -^k

| (\to a \times \to b) . (\to c \times \to d) | =

\to a = 2^i + 3^j +^k

\to b =^i -^j +^k

\to c =^i +^j +^k

\to d

\to a \times \to b = \to d \times \to b, \to d \cdot \to c = 8

\to d . \to b

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