Let A, B, C be three points whose position vectors respectively are

$\vec{a} = \hat{i} + 4 \hat{j} + 3 \hat{k}$

$\vec{b} = 2 \hat{i} + α \hat{j} + 4 \hat{k}, α \in R$

$\vec{c} = 3 \hat{i} - 2 \hat{j} + 5 \hat{k}$

If a is the smallest positive integer for which $\vec{a}, \overset{\leftarrow}{b}, \vec{c}$ are noncollinear, then the length of the median, in $Δ A B C,$ through A is:

Option 1 -

$\frac{\sqrt[]{82}}{2}$

Option 2 -

$\frac{\sqrt[]{62}}{2}$

Option 3 -

$\frac{\sqrt[]{69}}{2}$

Option 4 -

$\frac{\sqrt[]{66}}{2}$

2 Views | Posted a month ago

Asked by Shiksha User

1 Answer

Answered by

Vishal Baghel | Contributor-Level 10

a month ago

Correct Option - 3

Detailed Solution:

Mid point of BC is $\frac{1}{2} (5 \hat{i} + (α - 2) \hat{j} + 9 \hat{k})$ $\frac{}{}$

$\bar{A B} = \hat{i} + (α - 4) \hat{j} + \hat{k}$

$\bar{A C} = \hat{i} + (- 2 - α) \hat{j} + \hat{k}$

For = 1, $\bar{A B}$ and $\bar{A C}$ will be collinear. So for non collinearity

= 2

Mid point of BC is <math> <mrow> <mfrac> <mrow> <mn>1</mn> </mrow> <mrow> <mn>2</mn> </mrow> </mfrac> <mrow> <mo> (</mo> <mrow> <mn>5</mn> <mover accent="true"> <mi>i</mi> <mo>^</mo> </mover> <mo>+</mo> <mrow> <mo> (</mo> <mrow> <mi>α</mi> <mo>−</mo> <mn>2</mn> </mrow> <mo>)</mo> </mrow> <mover accent="true"> <mi>j</mi> <mo>^</mo> </mover> <mo>+</mo> <mn>9</mn> <mover accent="true"> <mi>k</mi> <mo>^</mo> </mover> </mrow> <mo>)</mo> </mrow> </mrow> </math> <math><mrow><mfrac><mrow></mrow><mrow></mrow></mfrac><mrow><mrow><mover accent="true"></mover><mrow><mrow></mrow></mrow><mover accent="true"></mover><mover accent="true"></mover></mrow></mrow></mrow></math><img > <math> <mrow> <mover accent="true"> <mrow> <mi>A</mi> <mi>B</mi> </mrow> <mo stretchy="true">¯</mo> </mover> <mo>=</mo> <mover accent="true"> <mi>i</mi> <mo>^</mo> </mover> <mo>+</mo> <mrow> <mo> (</mo> <mrow> <mi>α</mi> <mo>−</mo> <mn>4</mn> </mrow> <mo>)</mo> </mrow> <mover accent="true"> <mi>j</mi> <mo>^</mo> </mover> <mo>+</mo> <mover accent="true"> <mi>k</mi> <mo>^</mo> </mover> </mrow> </math> <math> <mrow> <mover accent="true"> <mrow> <mi>A</mi> <mi>C</mi> </mrow> <mo stretchy="true">¯</mo> </mover> <mo>=</mo> <mover accent="true"> <mi>i</mi> <mo>^</mo> </mover> <mo>+</mo> <mrow> <mo> (</mo> <mrow> <mo>−</mo> <mn>2</mn> <mo>−</mo> <mi>α</mi> </mrow> <mo>)</mo> </mrow> <mover accent="true"> <mi>j</mi> <mo>^</mo> </mover> <mo>+</mo> <mover accent="true"> <mi>k</mi> <mo>^</mo> </mover> </mrow> </math> For = 1, <math> <mrow> <mover accent="true"> <mrow> <mi>A</mi> <mi>B</mi> </mrow> <mo stretchy="true">¯</mo> </mover> </mrow> </math> <math><mrow><mover accent="true"><mrow></mrow></mover></mrow></math> and <math> <mrow> <mover accent="true"> <mrow> <mi>A</mi> <mi>C</mi> </mrow> <mo stretchy="true">¯</mo> </mover> </mrow> </math> <math><mrow><mover accent="true"><mrow></mrow></mover></mrow></math> will be collinear. So for non collinearity= 2

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