Let a- - b-k- and c-k- If d is a unit vector- such that a-d-0-bcd - then d is-

Given: nbsp;a=î ĵ , b=ĵ k̂ and c=k̂ î, d is a unit vector such that a·d=0=[bc nbsp;d nbsp;] Let d=d_1î+d_2ĵ+d_3k̂ a·d=d_1 d_2=0 ⇒ d_1=d_2 ⋯(i) And, [bc ...

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

Byju's Answer

Standard X

Mathematics

Section Formula

Let a=î-ĵ , b...

Question

Let $\to a =^i -^j, \to b =^j -^k$ and $\to c =^k -^i$ . If $\to d$ is a unit vector, such that $\to a \cdot \to d = 0 = [\to b \to c \to d]$ , then $\to d$ is:

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

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

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

No worries! We‘ve got your back. Try BYJU‘S free classes today!

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

No worries! We‘ve got your back. Try BYJU‘S free classes today!

\pm^k

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is A $\pm \frac{^i +^j - 2^k}{\sqrt{6}}$
Given: $\to a =^i -^j, \to b =^j -^k$ and $\to c =^k -^i$ , $\to d$ is a unit vector such that $\to a \cdot \to d = 0 = [\to b \to c \to d]$
Let $\to d = d_{1}^i + d_{2}^j + d_{3}^k$
$\to a \cdot \to d = d_{1} - d_{2} = 0$
$\Rightarrow d_{1} = d_{2} \dots (i)$

And, $[\to b \to c \to d] = 0$
$\Rightarrow ∣ ∣ ∣ ∣ \begin{matrix} 0 & 1 & - 1 - 1 & 0 & 1 d_{1} & d_{2} & d_{3} \end{matrix} ∣ ∣ ∣ ∣ = 0$
$\Rightarrow d_{3} + d_{1} + d_{2} = 0$
$\Rightarrow d_{3} = - 2 d_{1} \dots (i i)$

where, $\to d$ is a unit vector
$\Rightarrow \sqrt{(d_{1})^{2} + (d_{2})^{2} + (- 2 d_{1})^{2}} = 1$
$\Rightarrow d_{1} = \pm \frac{1}{\sqrt{6}}$
$∴ \to d = \pm \frac{^i +^j - 2^k}{\sqrt{6}}$

Suggest Corrections

Similar questions

Q. Let

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

. If

\to d

is a unit vector such that

\to a \cdot \to d = 0 = [\to b \to c \to d]

, then

\to d

is equal to:

L e t \to a =^i -^j, \to b =^j -^k, \to c =^k -^i .

\to d

is a unit vector such that

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

then

\to d

Q. Let

\to a = 2^i + 3^j -^k, \to b = -^i + 2^j - 4^k, \to c =^i +^j +^k, \to d =^i -^j -^k,

then

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

is equal to

Q. Let a vector

\to a

be coplanar with vectors

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

and

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

\to a

is perpendicular to

\to d = 3^i + 2^j + 6^k

, and

| \to a | = \sqrt{10} .

Then a possible value of

[\to a \to b \to c] + [\to a \to b \to d] + [\to a \to c \to d]

is equal to

Q. If

\to a = 2^i +^j + 3^k, \to b = 3^i + 2^j +^k, \to c =^i -^j - 4^k

and

\to d =^i + 2^j -^k,

then

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

Join BYJU'S Learning Program

Let →a=^i−^j,→b=^j−^k and →c=^k−^i. If →d is a unit vector, such that →a⋅→d=0=[→b→c→d], then →d is:

Let $\to a =^i -^j, \to b =^j -^k$ and $\to c =^k -^i$ . If $\to d$ is a unit vector, such that $\to a \cdot \to d = 0 = [\to b \to c \to d]$ , then $\to d$ is: