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<strong>Marking</strong> <strong>Scheme</strong><br />
331<br />
1<br />
1<br />
<strong>Marking</strong> scheme<br />
√<br />
√<br />
ITEM – 6(b)<br />
ITEM – 7(a)<br />
<strong>Marking</strong> scheme<br />
let, ⃗ ⃗ ⃗<br />
⃗ ⃗ ⃗<br />
Now, since, ⃗ and ⃗ are perpendicular to each other,<br />
then, we have<br />
⃗ ⃗<br />
( ⃗ ⃗) ( ⃗ ⃗)<br />
1<br />
1<br />
1<br />
1<br />
ITEM – 7(b)<br />
<strong>Marking</strong> scheme<br />
⃗⃗⃗⃗⃗⃗ ⃗<br />
⃗⃗⃗⃗⃗⃗ ⃗<br />
ratio ( ) 1<br />
⃗⃗⃗⃗⃗⃗⃗<br />
⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗<br />
⃗<br />
⃗<br />
⃗<br />
⃗<br />
1<br />
<strong>Marking</strong> scheme<br />
ITEM – 8(a)<br />
* +<br />
1<br />
( ) → ( ) 1<br />
<strong>Marking</strong> scheme<br />
Taking object point (<br />
* + * + * +<br />
ITEM – 8(b)<br />
) and expressing in matrix multiplication<br />
Identifying( ) ( ) 1<br />
1<br />
3
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