If vectors AB=3 hati-3 hatk and AC= hati-2 hatj+ hatk are the sides of a triangle ABC, then the length of the median AM, is

Question

If vectors AB=3 hati-3 hatk and AC= hati-2 hatj+ hatk are the sides of a triangle ABC, then the length of the median AM, is (A) √3 (B) √6 (C) 2√3 (D) 3√2

Tardigrade Admin · Accepted Answer

If AM is the.medium of Δ ABC then AB +AC=2 AM <img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/mathematics/b90fc27106e52cd5953106d151fa81e2-.jpeg" /> ⇒ 3 hati-3 hatk+ hati-2 hatj+ hatk=2 AM or 2 AM=4 hati-2 hatj-2 hatk ∴ AM=2 hati- hatj- hatk ∴ |AM| i.e., length AM=√4+1+1 =√6

Q. If vectors $A B = 3 \hat{i} - 3 \hat{k}$ and $A C = \hat{i} - 2 \hat{j} + \hat{k}$ are the sides of a triangle ABC, then the length of the median AM, is

$3$

$6$

$23$

$32$

If $A M$ is the.medium of $Δ A BC$ then
$A B + A C = 2 A M$

$\Rightarrow 3 \hat{i} - 3 \hat{k} + \hat{i} - 2 \hat{j} + \hat{k} = 2 A M$
or $2 A M = 4 \hat{i} - 2 \hat{j} - 2 \hat{k}$
$∴ A M = 2 \hat{i} - \hat{j} - \hat{k}$
$∴ ∣ ∣ A M ∣ ∣$ i.e., length $A M = 4 + 1 + 1$
$= 6$

Q. If vectors AB=3i^−3k^ and AC=i^−2j^​+k^ are the sides of a triangle ABC, then the length of the median AM, is

3​

6​

23​

32​

If AM is the.medium of ΔABC then AB+AC=2AM ⇒3i^−3k^+i^−2j^​+k^=2AM or 2AM=4i^−2j^​−2k^ ∴AM=2i^−j^​−k^ ∴∣∣​AM∣∣​ i.e., length AM=4+1+1​ =6​

Q. If vectors $A B = 3 \hat{i} - 3 \hat{k}$ and $A C = \hat{i} - 2 \hat{j} + \hat{k}$ are the sides of a triangle ABC, then the length of the median AM, is

$3$

$6$

$23$

$32$

If $A M$ is the.medium of $Δ A BC$ then
$A B + A C = 2 A M$

$\Rightarrow 3 \hat{i} - 3 \hat{k} + \hat{i} - 2 \hat{j} + \hat{k} = 2 A M$
or $2 A M = 4 \hat{i} - 2 \hat{j} - 2 \hat{k}$
$∴ A M = 2 \hat{i} - \hat{j} - \hat{k}$
$∴ ∣ ∣ A M ∣ ∣$ i.e., length $A M = 4 + 1 + 1$
$= 6$