If a- - - 2k- then - a- - -2-a-k-2-a-k-2-

⇒|(a×î)×ĵ|^2=|(a·ĵ)î (î·ĵ)a|^2=|î|^2=1 Similarly, ⇒|(a×ĵ)×k̂|^2=|(a·k̂)ĵ (k̂·ĵ)a|^2=| 2ĵ|^2=4 Similarly ⇒|(a×k̂)×î|^2=|(a·î)·k̂ (î·k̂)a|^2=|k̂|^2=1 Adding all a ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Physics

Subtraction of a Vector

If a= î+ ĵ- 2...

Question

If $\to a =^i +^j - 2^k,$ then ${∣ ∣ (\to a \times^i) \times^j ∣ ∣}^{2} + {∣ ∣ (\to a \times^j) \times^k ∣ ∣}^{2} + {∣ ∣ (\to a \times^k) \times^i ∣ ∣}^{2} =$

Open in App

Solution

$\Rightarrow | (\to a \times^i) \times^j |^{2} = | (\to a \cdot^j)^i - (^i \cdot^j) \to a |^{2} = |^i |^{2} = 1$
Similarly,
$\Rightarrow | (\to a \times^j) \times^k |^{2} = | (\to a \cdot^k)^j - (^k \cdot^j) \to a |^{2} = | - 2^j |^{2} = 4$
Similarly
$\Rightarrow | (\to a \times^k) \times^i |^{2} = | (\to a \cdot^i) \cdot^k - (^i \cdot^k) \to a |^{2} = |^k |^{2} = 1$
Adding all above three cases, We get
$\Rightarrow {∣ ∣ (\to a \times^i) \times^j ∣ ∣}^{2} + {∣ ∣ (\to a \times^j) \times^k ∣ ∣}^{2} + {∣ ∣ (\to a \times^k) \times^i ∣ ∣}^{2} = 6$

Suggest Corrections

Similar questions

Q. If

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

then the value of

{∣ ∣^i \times (\to a \times^i) ∣ ∣}^{2} + {∣ ∣^j \times (\to a \times^j) ∣ ∣}^{2} + {∣ ∣^k \times (\to a \times^k) ∣ ∣}^{2}

is equal to

Q. A vector of magnitude

\sqrt{2}

coplanar with the vectors

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

and

\to b =^i + 2^j +^k

and perpendicular to the vector

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

is
(1)

-^i -^k

(2)

^j -^k

(3)

^i -^j

(4)

^i +^k

Q. If

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

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

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

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

then

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

^i \times (^j \times^k) +^j \times (^k \times^i) +^k \times (^i \times^j) =

Q. If

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

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

and

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

, then

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

Join BYJU'S Learning Program

If →a=^i+^j−2^k, then ∣∣(→a×^i)×^j∣∣2+∣∣(→a×^j)×^k∣∣2+∣∣(→a×^k)×^i∣∣2=

If $\to a =^i +^j - 2^k,$ then ${∣ ∣ (\to a \times^i) \times^j ∣ ∣}^{2} + {∣ ∣ (\to a \times^j) \times^k ∣ ∣}^{2} + {∣ ∣ (\to a \times^k) \times^i ∣ ∣}^{2} =$