If {p}=2 {i}+{k}, {q}={i}+{j}+{k}, {r}=4 {i}-3 {j}+7 {k} and a vector {{m}} is such that {{m}} × {{q}}={{r}} × {{q}}, {{m}} · {{p}}=0, then {{m}}=.

Correct Answer is : −i^−8j^+2k^-{{i}}-8 {{j}}+2 {{k}}−i^−8j^+2k^

Correct Answer is : −i^−8j^+2k^-{{i}}-8 {{j}}+2 {{k}}−i^−8j^+2k^

Test Index

MHTCET 22 Apr 2025 Shift 2 Paper

Question : 125 of 150

Marks: +1, -0

\bar{p}=2 \hat{i}+\hat{k}, \bar{q}=\hat{i}+\hat{j}+\hat{k}, \bar{r}=4 \hat{i}-3 \hat{j}+7 \hat{k}

\overline{\mathrm{m}}

\overline{\mathrm{m}} \times \overline{\mathrm{q}}=\overline{\mathrm{r}} \times \overline{\mathrm{q}}, \overline{\mathrm{m}} \cdot \overline{\mathrm{p}}=0

\overline{\mathrm{m}}=\ldots

Solution:

Concept:

Assume

\overline{\mathrm{m}}=x\hat{\mathrm{i}}+y\hat{\mathrm{j}}+z\hat{\mathrm{k}}

and use both the cross product and dot product conditions to find

x

y

, and

z

Explanation:

Let

\overline{\mathrm{m}}=x\hat{\mathrm{i}}+y\hat{\mathrm{j}}+z\hat{\mathrm{k}}

Given

\overline{\mathrm{m}}\times\overline{\mathrm{q}}=\overline{\mathrm{r}}\times\overline{\mathrm{q}}

< br>\overline{\mathrm{m}}\times\overline{\mathrm{q}}= < br>\begin{vmatrix} < br>\hat{\mathrm{i}} & \hat{\mathrm{j}} & \hat{\mathrm{k}}\\ < br>x & y & z\\ < br>1 & 1 & 1 < br>\end{vmatrix} < br>=(y-z)\hat{\mathrm{i}}+(z-x)\hat{\mathrm{j}}+(x-y)\hat{\mathrm{k}} < br>

< br>\overline{\mathrm{r}}\times\overline{\mathrm{q}}= < br>\begin{vmatrix} < br>\hat{\mathrm{i}} & \hat{\mathrm{j}} & \hat{\mathrm{k}}\\ < br>4 & -3 & 7\\ < br>1 & 1 & 1 < br>\end{vmatrix} < br>=-10\hat{\mathrm{i}}+3\hat{\mathrm{j}}+7\hat{\mathrm{k}} < br>

Equating components gives:

y-z=-10

z-x=3

x-y=7

Also,

\overline{\mathrm{m}}\cdot\overline{\mathrm{p}}=0

and

\overline{\mathrm{p}}=2\hat{\mathrm{i}}+\hat{\mathrm{k}}

, so

2x+z=0

From

z-x=3

, we get

z=x+3

Substitute into

2x+z=0

2x+(x+3)=0

, hence

x=-1

Then

z=x+3=2

, and

y-z=-10

gives

y=-8

Thus,

\overline{\mathrm{m}}=-\hat{\mathrm{i}}-8\hat{\mathrm{j}}+2\hat{\mathrm{k}}

Answer:

\overline{\mathrm{m}}=-\hat{\mathrm{i}}-8\hat{\mathrm{j}}+2\hat{\mathrm{k}}

This matches option C.

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