{a}={i}-{j}, {b}={j}-{k}, {c}={k}-{i} then a unit vector {d} such that {a} · {d}=0=[{b} {c} {d}] is

Correct Answer is : ±(i^+j^−2k^6)±≤ft({{{i}}+{{j}}-2 {{k}}}{{6}})±(6i^+j^−2k^)

±≤ft({{{i}}+{{j}}+3 {{k}}}{{11}})

±≤ft({{{i}}+{{j}}+{{k}}}{{3}})

±≤ft({{{i}}+{{j}}-2 {{k}}}{{6}})

Correct Answer is : ±(i^+j^−2k^6)±≤ft({{{i}}+{{j}}-2 {{k}}}{{6}})±(6i^+j^−2k^)

Test Index

MHTCET 22 Apr 2025 Shift 1 Paper

Question : 107 of 150

Marks: +1, -0

\bar{a}=\hat{i}-\hat{j}, \bar{b}=\hat{j}-\hat{k}, \bar{c}=\hat{k}-\hat{i}

\bar{d}

\overline{a} \cdot \overline{d}=0=[\overline{b} \overline{c} \overline{d}]

$\pm\left(\frac{\hat{\mathbf{i}}+\hat{\mathbf{j}}+3 \hat{\mathbf{k}}}{\sqrt{11}}\right)$
$\pm\left(\frac{-\hat{\mathrm{j}}+\hat{\mathrm{k}}}{\sqrt{2}}\right)$
$\pm\left(\frac{\hat{\mathrm{i}}+\hat{\mathrm{j}}+\hat{\mathrm{k}}}{\sqrt{3}}\right)$
$\pm\left(\frac{\hat{\mathrm{i}}+\hat{\mathrm{j}}-2 \hat{\mathrm{k}}}{\sqrt{6}}\right)$

Solution:

Concept:

A unit vector must satisfy both

\bar{a} \cdot \bar{d}=0

and the coplanarity condition

[\bar{b}\,\bar{c}\,\bar{d}]=0

Explanation:

Let

\bar{d}=p\hat{i}+q\hat{j}+r\hat{k}

, where

p,q,r\in\mathbb{R}

Since

\bar{a}\cdot\bar{d}=0

, we get:

(\hat{i}-\hat{j})\cdot(p\hat{i}+q\hat{j}+r\hat{k})=0

p-q=0 \Rightarrow p=q

Now,

[\bar{b}\,\bar{c}\,\bar{d}]=0

means

\bar{b},\bar{c},\bar{d}

are coplanar. Therefore:

\begin{vmatrix} 0 & 1 & -1\\ -1 & 0 & 1\\ p & q & r \end{vmatrix}=0

Expanding the determinant:

-(-r-p)-(-q)=0

p+q+r=0

Using

p=q

, we get

2p+r=0 \Rightarrow r=-2p

Thus,

\bar{d}=p\hat{i}+p\hat{j}-2p\hat{k}=p(\hat{i}+\hat{j}-2\hat{k})

Since

\bar{d}

is a unit vector,

|p|\sqrt{1+1+4}=1

gives

p=\pm\frac{1}{\sqrt{6}}

Therefore:

\bar{d}=\pm\frac{\hat{i}+\hat{j}-2\hat{k}}{\sqrt{6}}

This matches option D.

Answer:

Option D:

\pm\left(\frac{\hat{i}+\hat{j}-2\hat{k}}{\sqrt{6}}\right)

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