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Vector Algebra

Section: Mathematics

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Question : 29 of 38

Marks: +1, -0

The vector(s) which is/are coplanar with vectors

\hat{i} + \hat{j} + \widehat{2k}

\hat{i} + \widehat{2j} + \hat{k}

, and perpendicular to the vector

\hat{i} + \hat{j} + \hat{k}

[JEE Adv 2011 P1]

$\hat{j} - \hat{k}$
$-\hat{i} + \hat{j}$
$\hat{i} - \hat{j}$ $\hat{i} - \hat{j}$
$-\hat{j} + \hat{k}$

Solution:

Let

\vec{a}

=

\hat{i} + \hat{j} + \widehat{2k}

,

\vec{b}

=

\hat{i} + \widehat{2j} + \hat{k}

and

\vec{c}

=

\hat{i} + \hat{j} + \hat{k}

Any vector in the plane of

\hat{i} + \hat{j} + \widehat{2k}

and

\hat{i} + \widehat{2j} + \hat{k}

is given by

\vec{r}

=

\lambda \vec{a} + \mu \vec{b}

=

\lambda(\hat{i}+\hat{j}+\widehat{2k}) + \mu(\hat{i}+\widehat{2j}+\hat{k})

=

(\lambda+\mu)\hat{i} + (\lambda+2\mu)\hat{j} + (2\lambda+\mu)\hat{k}

Also ,

\vec{r} \cdot \vec{c}

= 0

⇒ (λ + µ) . 1 + (λ + 2µ) . 1 + (2λ + µ) . 1 = 0

⇒ 4λ + 4µ = 0

⇒ λ + µ = 0

⇒

\begin{bmatrix} \vec{r} & \vec{a} & \vec{b} \end{bmatrix}

= 0

So, vectors

\hat{j} - \hat{k}

and

-\hat{j} + \hat{k}

satisfy this.

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