The unit vector which is orthogonal to the vector 3-2-6k- is coplanar with vectors 2-6k- and -k- is

(a⃗) × (b⃗×c⃗) is cross product of two vectors (a⃗) and (b⃗×c⃗). So this vector is orthogonal to a⃗ and coplanar with b⃗ and c⃗. So the vector orthogonal to 3î+ ...

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Standard XII

Mathematics

Vector Triple Product

The unit vect...

Question

The unit vector which is orthogonal to the vector $3^i + 2^j + 6^k$ is coplanar with vectors $2^i +^j + 6^k a n d^i -^j -^k$ is

\frac{2^i - 6^j + k}{\sqrt{41}}

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\frac{2^i - 3^j}{\sqrt{13}}

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\frac{3^j -^k}{\sqrt{10}}

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\frac{4^i + 3^j + 3 k}{\sqrt{34}}

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Solution

The correct option is C $\frac{3^j -^k}{\sqrt{10}}$
$(\to a) \times (\to b \times \to c)$ is cross product of two vectors $(\to a) a n d (\to b \times \to c)$ .
So this vector is orthogonal to $\to a$ and coplanar with $\to b a n d \to c$ .
So the vector orthogonal to $3^i + 2^j + 6^k$ and coplanar with $2^i +^j +^k a n d^i -^j -^k$ will be given by $(3^i + 2^j + 6^k) \times ((2^i +^j +^k) \times (^i -^j -^k))$
Vector triple product
Let $3^i + 2^j + 6^k = \to a$
$2^i +^j +^k = \to b$
$^i -^j -^k = \to c$
So we have to calculate $\to a \times (\to b \times \to c)$
$\to a \times (\to b \times \to c) = (\to a . \to c) \to b - (\to a . \to b) \to c$
$(\to a . \to c) = (3^i + 2^j + 6^k) . (^i -^j -^k)$
= 3 - 2 + 6
= 7
$(\to a . \to b) = (3^i + 2^j + 6^k) . (2^i +^j +^k)$ = 6 + 2 + 6
= 14
So $\to a \times (\to b \times \to c) = 7 (2^i +^j +^k) - 14 (^i -^j -^k)$
$= 21^j - 7^k$
So the unit vector for this vector will be obtained by dividing it by its magnitude
Magnitude of $21^j - 7^k = \sqrt{(21)^{2} + (7)^{2}}$
$= \sqrt{490}$
$= 7 \sqrt{10}$
So unit vector of $21^j - 7^k$
$= \frac{21^j - 7^k}{7 \sqrt{10}}$
$= \frac{7 (3^j - 7^k)}{7 \sqrt{10}}$
$= \frac{3^j -^k}{\sqrt{10}} .$

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The unit vector which is orthogonal to the vector 3^i+2^j+6^k is coplanar with vectors 2^i+^j+6^k and ^i−^j−^k is

The unit vector which is orthogonal to the vector $3^i + 2^j + 6^k$ is coplanar with vectors $2^i +^j + 6^k a n d^i -^j -^k$ is